I don't understand how the time complexity for this algorithm is calculated











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int j=0;

for (int i=0; i<N; i++) 

{

    while ( (j<N-1) && (A[i]-A[j] > D) )

        j++;

    if (A[i]-A[j] == D) 

        return 1; 

}


This code is said to have the time complexity of O(n), but I don't really get it. The inner loop is executed N times and the outer should be also N times? Is it maybe because of the j = 0; outside the loop that is making it only run N times?



But even if it would only run N times in the inner loop, the if statment check should be done also N times, which should bring the total time complexity to O(n^2)?






c algorithm loops time-complexity big-o







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  • 5




    Look at it this way: j++ won't be executed more than N-1 times. It's not set to 0 at each outer loop iteration start.
    – algrid
    3 hours ago






  • 1




    The inner loop is not repeatedly executed N times. That will only happen once. Once (j<N-1) is false that loop will never be entered again.
    – WhozCraig
    3 hours ago










  • Possible duplicate of How to find time complexity of an algorithm
    – cirrusio
    3 hours ago















up vote
8
down vote

favorite
1












int j=0;

for (int i=0; i<N; i++) 

{

    while ( (j<N-1) && (A[i]-A[j] > D) )

        j++;

    if (A[i]-A[j] == D) 

        return 1; 

}


This code is said to have the time complexity of O(n), but I don't really get it. The inner loop is executed N times and the outer should be also N times? Is it maybe because of the j = 0; outside the loop that is making it only run N times?



But even if it would only run N times in the inner loop, the if statment check should be done also N times, which should bring the total time complexity to O(n^2)?






c algorithm loops time-complexity big-o







share|improve this question









New contributor




lomo133 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 5




    Look at it this way: j++ won't be executed more than N-1 times. It's not set to 0 at each outer loop iteration start.
    – algrid
    3 hours ago






  • 1




    The inner loop is not repeatedly executed N times. That will only happen once. Once (j<N-1) is false that loop will never be entered again.
    – WhozCraig
    3 hours ago










  • Possible duplicate of How to find time complexity of an algorithm
    – cirrusio
    3 hours ago













up vote
8
down vote

favorite
1









up vote
8
down vote

favorite
1






1





int j=0;

for (int i=0; i<N; i++) 

{

    while ( (j<N-1) && (A[i]-A[j] > D) )

        j++;

    if (A[i]-A[j] == D) 

        return 1; 

}


This code is said to have the time complexity of O(n), but I don't really get it. The inner loop is executed N times and the outer should be also N times? Is it maybe because of the j = 0; outside the loop that is making it only run N times?



But even if it would only run N times in the inner loop, the if statment check should be done also N times, which should bring the total time complexity to O(n^2)?






c algorithm loops time-complexity big-o







share|improve this question









New contributor




lomo133 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct. 











int j=0;

for (int i=0; i<N; i++) 

{

    while ( (j<N-1) && (A[i]-A[j] > D) )

        j++;

    if (A[i]-A[j] == D) 

        return 1; 

}


This code is said to have the time complexity of O(n), but I don't really get it. The inner loop is executed N times and the outer should be also N times? Is it maybe because of the j = 0; outside the loop that is making it only run N times?



But even if it would only run N times in the inner loop, the if statment check should be done also N times, which should bring the total time complexity to O(n^2)?






c algorithm loops time-complexity big-o




c algorithm loops time-complexity big-o






share|improve this question









New contributor




lomo133 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




lomo133 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 4 hours ago









Some programmer dude

293k24243403




293k24243403






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asked 4 hours ago









lomo133

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413




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lomo133 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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Check out our Code of Conduct.








  • 5




    Look at it this way: j++ won't be executed more than N-1 times. It's not set to 0 at each outer loop iteration start.
    – algrid
    3 hours ago






  • 1




    The inner loop is not repeatedly executed N times. That will only happen once. Once (j<N-1) is false that loop will never be entered again.
    – WhozCraig
    3 hours ago










  • Possible duplicate of How to find time complexity of an algorithm
    – cirrusio
    3 hours ago














  • 5




    Look at it this way: j++ won't be executed more than N-1 times. It's not set to 0 at each outer loop iteration start.
    – algrid
    3 hours ago






  • 1




    The inner loop is not repeatedly executed N times. That will only happen once. Once (j<N-1) is false that loop will never be entered again.
    – WhozCraig
    3 hours ago










  • Possible duplicate of How to find time complexity of an algorithm
    – cirrusio
    3 hours ago








5




5




Look at it this way: j++ won't be executed more than N-1 times. It's not set to 0 at each outer loop iteration start.
– algrid
3 hours ago




Look at it this way: j++ won't be executed more than N-1 times. It's not set to 0 at each outer loop iteration start.
– algrid
3 hours ago




1




1




The inner loop is not repeatedly executed N times. That will only happen once. Once (j<N-1) is false that loop will never be entered again.
– WhozCraig
3 hours ago




The inner loop is not repeatedly executed N times. That will only happen once. Once (j<N-1) is false that loop will never be entered again.
– WhozCraig
3 hours ago












Possible duplicate of How to find time complexity of an algorithm
– cirrusio
3 hours ago




Possible duplicate of How to find time complexity of an algorithm
– cirrusio
3 hours ago












1 Answer
1






active

oldest

votes

















up vote
10
down vote













The reason why this is O(n) is because j is not set back to 0 in the body of the for loop.



Indeed if we take a look at the body of the for loop, we see:



while ( (j<N-1) && (A[i]-A[j] > D) )

    j++;


That thus means that j++ is done at most n-1 times, since if j succeeds N-1 times, then the first constraint fails.



If we take a look at the entire for loop, we see:



int j=0;

for (int i=0; i<N; i++) {

    while ( (j<N-1) && (A[i]-A[j] > D) )

        j++;

    if (A[i]-A[j] == D) 

        return 1;

}


It is clear that the body of the for loop is repeated n times, since we set i to i=0, and stop when i >= N, and each iteration we increment i.



Now depending on the values in A we will or will not increment j (multiple times) in the body of the for loop. But regardless how many times it is done in a single iteration, at the end of the for loop, j++ is done at most n times, for the reason we mentioned above.



The condition in the while loop is executed O(n) (well at most 2×n-1 times to be precise) times as well: it is executed once each time we enter the body of the for loop, and each time after we execute a j++ command, but since both are O(n), this is done at most O(n+n) thus O(n) times.



The if condition in the for loop executed n times: once per iteration of the for loop, so again O(n).



So this indeed means that all "basic instructions" (j++, i = 0, j = 0, j < N-1, etc.) are all done either a constant number of times O(1), or a linear number of times O(n), hence the algorithm is O(n).






share|improve this answer



















  • 1




    No, but you can still say "at most n times." Suffice it to say that I've been saying "the loop executes n times" my whole career, and I don't see any compelling reason to complicate that simple phrase.
    – Robert Harvey♦
    3 hours ago






  • 1




    I consider your last example correct usage.
    – Robert Harvey♦
    2 hours ago






  • 2




    @robertharvey: O(n) describes the asymptotic behaviour of a function. It is orthogonal to the purpose of that function, which is outside the realm of mathematics. Now, if I have a loop in my program, I can certainly say that the loop condition will be true g(X) times, where g(X) is a function mapping algorithm input X to integers. If I can also demonstrate that g(X) is in O(f(|X|)) for some function f which maps integers to integers, then I am completely justified in saying the loop executes O(f(N)) times where N is the size of the problem. Why shouldn't I?
    – rici
    2 hours ago






  • 1




    Although I like the discussion, I modified it to more "rigorous" notation. Something that is also noteworthy is that the error of approximative sequences is typically expressed in big oh as well, since here one typically has not enough information at all to specify the exact error (or at least not without exhaustive analysis, that might not be the scope of the paper), as is mentioned in the wiki article. Somehow I forgot about that (been away from academia for too long I guess) en.wikipedia.org/wiki/Big_O_notation#Infinitesimal_asymptotics
    – Willem Van Onsem
    2 hours ago






  • 1




    @Robert: perhaps you disagree but I don't like saying that something happens at most n times when it can happen 2n times. OP asked a question using Big-O notation so responding with Big-O notation is not tossing a completely unfamiliar concept at them. They even identified the issue of counting executions of the test, but got the asymptote wrong. Anyway, we're obviously not going to get any further here right now and I've got a non-CS event to go to. So take care...
    – rici
    2 hours ago











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1 Answer
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1 Answer
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active

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up vote
10
down vote













The reason why this is O(n) is because j is not set back to 0 in the body of the for loop.



Indeed if we take a look at the body of the for loop, we see:



while ( (j<N-1) && (A[i]-A[j] > D) )

    j++;


That thus means that j++ is done at most n-1 times, since if j succeeds N-1 times, then the first constraint fails.



If we take a look at the entire for loop, we see:



int j=0;

for (int i=0; i<N; i++) {

    while ( (j<N-1) && (A[i]-A[j] > D) )

        j++;

    if (A[i]-A[j] == D) 

        return 1;

}


It is clear that the body of the for loop is repeated n times, since we set i to i=0, and stop when i >= N, and each iteration we increment i.



Now depending on the values in A we will or will not increment j (multiple times) in the body of the for loop. But regardless how many times it is done in a single iteration, at the end of the for loop, j++ is done at most n times, for the reason we mentioned above.



The condition in the while loop is executed O(n) (well at most 2×n-1 times to be precise) times as well: it is executed once each time we enter the body of the for loop, and each time after we execute a j++ command, but since both are O(n), this is done at most O(n+n) thus O(n) times.



The if condition in the for loop executed n times: once per iteration of the for loop, so again O(n).



So this indeed means that all "basic instructions" (j++, i = 0, j = 0, j < N-1, etc.) are all done either a constant number of times O(1), or a linear number of times O(n), hence the algorithm is O(n).






share|improve this answer



















  • 1




    No, but you can still say "at most n times." Suffice it to say that I've been saying "the loop executes n times" my whole career, and I don't see any compelling reason to complicate that simple phrase.
    – Robert Harvey♦
    3 hours ago






  • 1




    I consider your last example correct usage.
    – Robert Harvey♦
    2 hours ago






  • 2




    @robertharvey: O(n) describes the asymptotic behaviour of a function. It is orthogonal to the purpose of that function, which is outside the realm of mathematics. Now, if I have a loop in my program, I can certainly say that the loop condition will be true g(X) times, where g(X) is a function mapping algorithm input X to integers. If I can also demonstrate that g(X) is in O(f(|X|)) for some function f which maps integers to integers, then I am completely justified in saying the loop executes O(f(N)) times where N is the size of the problem. Why shouldn't I?
    – rici
    2 hours ago






  • 1




    Although I like the discussion, I modified it to more "rigorous" notation. Something that is also noteworthy is that the error of approximative sequences is typically expressed in big oh as well, since here one typically has not enough information at all to specify the exact error (or at least not without exhaustive analysis, that might not be the scope of the paper), as is mentioned in the wiki article. Somehow I forgot about that (been away from academia for too long I guess) en.wikipedia.org/wiki/Big_O_notation#Infinitesimal_asymptotics
    – Willem Van Onsem
    2 hours ago






  • 1




    @Robert: perhaps you disagree but I don't like saying that something happens at most n times when it can happen 2n times. OP asked a question using Big-O notation so responding with Big-O notation is not tossing a completely unfamiliar concept at them. They even identified the issue of counting executions of the test, but got the asymptote wrong. Anyway, we're obviously not going to get any further here right now and I've got a non-CS event to go to. So take care...
    – rici
    2 hours ago















up vote
10
down vote













The reason why this is O(n) is because j is not set back to 0 in the body of the for loop.



Indeed if we take a look at the body of the for loop, we see:



while ( (j<N-1) && (A[i]-A[j] > D) )

    j++;


That thus means that j++ is done at most n-1 times, since if j succeeds N-1 times, then the first constraint fails.



If we take a look at the entire for loop, we see:



int j=0;

for (int i=0; i<N; i++) {

    while ( (j<N-1) && (A[i]-A[j] > D) )

        j++;

    if (A[i]-A[j] == D) 

        return 1;

}


It is clear that the body of the for loop is repeated n times, since we set i to i=0, and stop when i >= N, and each iteration we increment i.



Now depending on the values in A we will or will not increment j (multiple times) in the body of the for loop. But regardless how many times it is done in a single iteration, at the end of the for loop, j++ is done at most n times, for the reason we mentioned above.



The condition in the while loop is executed O(n) (well at most 2×n-1 times to be precise) times as well: it is executed once each time we enter the body of the for loop, and each time after we execute a j++ command, but since both are O(n), this is done at most O(n+n) thus O(n) times.



The if condition in the for loop executed n times: once per iteration of the for loop, so again O(n).



So this indeed means that all "basic instructions" (j++, i = 0, j = 0, j < N-1, etc.) are all done either a constant number of times O(1), or a linear number of times O(n), hence the algorithm is O(n).






share|improve this answer



















  • 1




    No, but you can still say "at most n times." Suffice it to say that I've been saying "the loop executes n times" my whole career, and I don't see any compelling reason to complicate that simple phrase.
    – Robert Harvey♦
    3 hours ago






  • 1




    I consider your last example correct usage.
    – Robert Harvey♦
    2 hours ago






  • 2




    @robertharvey: O(n) describes the asymptotic behaviour of a function. It is orthogonal to the purpose of that function, which is outside the realm of mathematics. Now, if I have a loop in my program, I can certainly say that the loop condition will be true g(X) times, where g(X) is a function mapping algorithm input X to integers. If I can also demonstrate that g(X) is in O(f(|X|)) for some function f which maps integers to integers, then I am completely justified in saying the loop executes O(f(N)) times where N is the size of the problem. Why shouldn't I?
    – rici
    2 hours ago






  • 1




    Although I like the discussion, I modified it to more "rigorous" notation. Something that is also noteworthy is that the error of approximative sequences is typically expressed in big oh as well, since here one typically has not enough information at all to specify the exact error (or at least not without exhaustive analysis, that might not be the scope of the paper), as is mentioned in the wiki article. Somehow I forgot about that (been away from academia for too long I guess) en.wikipedia.org/wiki/Big_O_notation#Infinitesimal_asymptotics
    – Willem Van Onsem
    2 hours ago






  • 1




    @Robert: perhaps you disagree but I don't like saying that something happens at most n times when it can happen 2n times. OP asked a question using Big-O notation so responding with Big-O notation is not tossing a completely unfamiliar concept at them. They even identified the issue of counting executions of the test, but got the asymptote wrong. Anyway, we're obviously not going to get any further here right now and I've got a non-CS event to go to. So take care...
    – rici
    2 hours ago













up vote
10
down vote










up vote
10
down vote









The reason why this is O(n) is because j is not set back to 0 in the body of the for loop.



Indeed if we take a look at the body of the for loop, we see:



while ( (j<N-1) && (A[i]-A[j] > D) )

    j++;


That thus means that j++ is done at most n-1 times, since if j succeeds N-1 times, then the first constraint fails.



If we take a look at the entire for loop, we see:



int j=0;

for (int i=0; i<N; i++) {

    while ( (j<N-1) && (A[i]-A[j] > D) )

        j++;

    if (A[i]-A[j] == D) 

        return 1;

}


It is clear that the body of the for loop is repeated n times, since we set i to i=0, and stop when i >= N, and each iteration we increment i.



Now depending on the values in A we will or will not increment j (multiple times) in the body of the for loop. But regardless how many times it is done in a single iteration, at the end of the for loop, j++ is done at most n times, for the reason we mentioned above.



The condition in the while loop is executed O(n) (well at most 2×n-1 times to be precise) times as well: it is executed once each time we enter the body of the for loop, and each time after we execute a j++ command, but since both are O(n), this is done at most O(n+n) thus O(n) times.



The if condition in the for loop executed n times: once per iteration of the for loop, so again O(n).



So this indeed means that all "basic instructions" (j++, i = 0, j = 0, j < N-1, etc.) are all done either a constant number of times O(1), or a linear number of times O(n), hence the algorithm is O(n).






share|improve this answer














The reason why this is O(n) is because j is not set back to 0 in the body of the for loop.



Indeed if we take a look at the body of the for loop, we see:



while ( (j<N-1) && (A[i]-A[j] > D) )

    j++;


That thus means that j++ is done at most n-1 times, since if j succeeds N-1 times, then the first constraint fails.



If we take a look at the entire for loop, we see:



int j=0;

for (int i=0; i<N; i++) {

    while ( (j<N-1) && (A[i]-A[j] > D) )

        j++;

    if (A[i]-A[j] == D) 

        return 1;

}


It is clear that the body of the for loop is repeated n times, since we set i to i=0, and stop when i >= N, and each iteration we increment i.



Now depending on the values in A we will or will not increment j (multiple times) in the body of the for loop. But regardless how many times it is done in a single iteration, at the end of the for loop, j++ is done at most n times, for the reason we mentioned above.



The condition in the while loop is executed O(n) (well at most 2×n-1 times to be precise) times as well: it is executed once each time we enter the body of the for loop, and each time after we execute a j++ command, but since both are O(n), this is done at most O(n+n) thus O(n) times.



The if condition in the for loop executed n times: once per iteration of the for loop, so again O(n).



So this indeed means that all "basic instructions" (j++, i = 0, j = 0, j < N-1, etc.) are all done either a constant number of times O(1), or a linear number of times O(n), hence the algorithm is O(n).







share|improve this answer














share|improve this answer



share|improve this answer








edited 2 hours ago

























answered 3 hours ago









Willem Van Onsem

142k16134226




142k16134226








  • 1




    No, but you can still say "at most n times." Suffice it to say that I've been saying "the loop executes n times" my whole career, and I don't see any compelling reason to complicate that simple phrase.
    – Robert Harvey♦
    3 hours ago






  • 1




    I consider your last example correct usage.
    – Robert Harvey♦
    2 hours ago






  • 2




    @robertharvey: O(n) describes the asymptotic behaviour of a function. It is orthogonal to the purpose of that function, which is outside the realm of mathematics. Now, if I have a loop in my program, I can certainly say that the loop condition will be true g(X) times, where g(X) is a function mapping algorithm input X to integers. If I can also demonstrate that g(X) is in O(f(|X|)) for some function f which maps integers to integers, then I am completely justified in saying the loop executes O(f(N)) times where N is the size of the problem. Why shouldn't I?
    – rici
    2 hours ago






  • 1




    Although I like the discussion, I modified it to more "rigorous" notation. Something that is also noteworthy is that the error of approximative sequences is typically expressed in big oh as well, since here one typically has not enough information at all to specify the exact error (or at least not without exhaustive analysis, that might not be the scope of the paper), as is mentioned in the wiki article. Somehow I forgot about that (been away from academia for too long I guess) en.wikipedia.org/wiki/Big_O_notation#Infinitesimal_asymptotics
    – Willem Van Onsem
    2 hours ago






  • 1




    @Robert: perhaps you disagree but I don't like saying that something happens at most n times when it can happen 2n times. OP asked a question using Big-O notation so responding with Big-O notation is not tossing a completely unfamiliar concept at them. They even identified the issue of counting executions of the test, but got the asymptote wrong. Anyway, we're obviously not going to get any further here right now and I've got a non-CS event to go to. So take care...
    – rici
    2 hours ago














  • 1




    No, but you can still say "at most n times." Suffice it to say that I've been saying "the loop executes n times" my whole career, and I don't see any compelling reason to complicate that simple phrase.
    – Robert Harvey♦
    3 hours ago






  • 1




    I consider your last example correct usage.
    – Robert Harvey♦
    2 hours ago






  • 2




    @robertharvey: O(n) describes the asymptotic behaviour of a function. It is orthogonal to the purpose of that function, which is outside the realm of mathematics. Now, if I have a loop in my program, I can certainly say that the loop condition will be true g(X) times, where g(X) is a function mapping algorithm input X to integers. If I can also demonstrate that g(X) is in O(f(|X|)) for some function f which maps integers to integers, then I am completely justified in saying the loop executes O(f(N)) times where N is the size of the problem. Why shouldn't I?
    – rici
    2 hours ago






  • 1




    Although I like the discussion, I modified it to more "rigorous" notation. Something that is also noteworthy is that the error of approximative sequences is typically expressed in big oh as well, since here one typically has not enough information at all to specify the exact error (or at least not without exhaustive analysis, that might not be the scope of the paper), as is mentioned in the wiki article. Somehow I forgot about that (been away from academia for too long I guess) en.wikipedia.org/wiki/Big_O_notation#Infinitesimal_asymptotics
    – Willem Van Onsem
    2 hours ago






  • 1




    @Robert: perhaps you disagree but I don't like saying that something happens at most n times when it can happen 2n times. OP asked a question using Big-O notation so responding with Big-O notation is not tossing a completely unfamiliar concept at them. They even identified the issue of counting executions of the test, but got the asymptote wrong. Anyway, we're obviously not going to get any further here right now and I've got a non-CS event to go to. So take care...
    – rici
    2 hours ago








1




1




No, but you can still say "at most n times." Suffice it to say that I've been saying "the loop executes n times" my whole career, and I don't see any compelling reason to complicate that simple phrase.
– Robert Harvey♦
3 hours ago




No, but you can still say "at most n times." Suffice it to say that I've been saying "the loop executes n times" my whole career, and I don't see any compelling reason to complicate that simple phrase.
– Robert Harvey♦
3 hours ago




1




1




I consider your last example correct usage.
– Robert Harvey♦
2 hours ago




I consider your last example correct usage.
– Robert Harvey♦
2 hours ago




2




2




@robertharvey: O(n) describes the asymptotic behaviour of a function. It is orthogonal to the purpose of that function, which is outside the realm of mathematics. Now, if I have a loop in my program, I can certainly say that the loop condition will be true g(X) times, where g(X) is a function mapping algorithm input X to integers. If I can also demonstrate that g(X) is in O(f(|X|)) for some function f which maps integers to integers, then I am completely justified in saying the loop executes O(f(N)) times where N is the size of the problem. Why shouldn't I?
– rici
2 hours ago




@robertharvey: O(n) describes the asymptotic behaviour of a function. It is orthogonal to the purpose of that function, which is outside the realm of mathematics. Now, if I have a loop in my program, I can certainly say that the loop condition will be true g(X) times, where g(X) is a function mapping algorithm input X to integers. If I can also demonstrate that g(X) is in O(f(|X|)) for some function f which maps integers to integers, then I am completely justified in saying the loop executes O(f(N)) times where N is the size of the problem. Why shouldn't I?
– rici
2 hours ago




1




1




Although I like the discussion, I modified it to more "rigorous" notation. Something that is also noteworthy is that the error of approximative sequences is typically expressed in big oh as well, since here one typically has not enough information at all to specify the exact error (or at least not without exhaustive analysis, that might not be the scope of the paper), as is mentioned in the wiki article. Somehow I forgot about that (been away from academia for too long I guess) en.wikipedia.org/wiki/Big_O_notation#Infinitesimal_asymptotics
– Willem Van Onsem
2 hours ago




Although I like the discussion, I modified it to more "rigorous" notation. Something that is also noteworthy is that the error of approximative sequences is typically expressed in big oh as well, since here one typically has not enough information at all to specify the exact error (or at least not without exhaustive analysis, that might not be the scope of the paper), as is mentioned in the wiki article. Somehow I forgot about that (been away from academia for too long I guess) en.wikipedia.org/wiki/Big_O_notation#Infinitesimal_asymptotics
– Willem Van Onsem
2 hours ago




1




1




@Robert: perhaps you disagree but I don't like saying that something happens at most n times when it can happen 2n times. OP asked a question using Big-O notation so responding with Big-O notation is not tossing a completely unfamiliar concept at them. They even identified the issue of counting executions of the test, but got the asymptote wrong. Anyway, we're obviously not going to get any further here right now and I've got a non-CS event to go to. So take care...
– rici
2 hours ago




@Robert: perhaps you disagree but I don't like saying that something happens at most n times when it can happen 2n times. OP asked a question using Big-O notation so responding with Big-O notation is not tossing a completely unfamiliar concept at them. They even identified the issue of counting executions of the test, but got the asymptote wrong. Anyway, we're obviously not going to get any further here right now and I've got a non-CS event to go to. So take care...
– rici
2 hours ago










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