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I have a method that finds 3 numbers in an array that add up to a desired number.

public static void threeSum(int[] arr, int sum) {
    quicksort(arr, 0, arr.length - 1);
    for (int i = 0; i < arr.length - 2; i++) {
        for (int j = 1; j < arr.length - 1; j++) {
            for (int k = arr.length - 1; k > j; k--) {
                if ((arr[i] + arr[j] + arr[k]) == sum) {
                    System.out.println(Integer.toString(i) + "+" + Integer.toString(j) + "+" + Integer.toString(k) + "="  + sum);

I'm not sure about the big O of this method. I have a hard time wrapping my head around this right now. My guess is O(n2) or O(n2logn). But these are complete guesses. I can't prove this. Could someone help me wrap my head around this?

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up vote 7 down vote accepted

The running time is O(n3). You have three nested loops:

  • Approximately n iterations of i
  • For each i, approximately n iterations of j
  • For each (i, j) pair, up to about n iterations of k. On average, it take just 0.5 * n iterations of k, but the constant factor doesn't matter.

Quicksort is generally O(n log n). So the total is O(n log n + 0.5 n × n × n), which is O(n3).

Since you took the trouble to sort the array, the innermost loop could be replaced by a binary search, which is O(log n). Therefore, an improved algorithm could work in O(n2 log n).

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Oh wow Binary search within the inner loop! Never though of that thank you! – Liondancer Apr 3 '14 at 7:33

This won't change the Big O but the second loop can start at j = i + 1.

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Are you looking for three distinct numbers, or can i, j, and k be the same? The loop limits suggest that you expect the numbers to be distinct. However, you don't actually check that the results are distinct.

To be consistent, you should either:

  • Make all three loops consider entries from the entire array, or
  • Discard solutions where i = j or where j = k or where i = k.
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I figured out the big O of my program. However you made a really good point about the values of the numbers. I am going to look into this! – Liondancer Apr 3 '14 at 7:25
O(n^3) for those wondering – Liondancer Apr 3 '14 at 7:33

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