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    <p><strong>O(N) time complexity and O(1) space complexity</strong>. This is the optimal solution I know:</p> <pre><code>#include &lt;stdio.h&gt; #include &lt;limits.h&gt; int get_max_sum(int* array, int len, int* start, int* end) { int max_sum = INT_MIN, sum = 0, i; int tmp_start = 0; for(i = 0; i != len; ++i) { sum += array[i]; // if the sum is equal, choose the one with more elements if(sum &gt; max_sum || (sum == max_sum &amp;&amp; (end - start) &lt; (i - tmp_start))) { max_sum = sum; *start = tmp_start; *end = i; } if(sum &lt; 0) { sum = 0; tmp_start = i + 1; } } return max_sum; } </code></pre> <p>Here are some test cases:</p> <pre><code>int main(int argc, char **argv) { int arr1[] = {5, 1, -7, 3, 7}; int arr2[] = {1}; int arr3[] = {-1, -7, -3, -7}; int arr4[] = {5, 1, -7, 2, 2, 2}; int start, end, sum; sum = get_max_sum(arr1, 5, &amp;start, &amp;end); printf("sum: %d, start: %d, end: %d\n", sum, start, end); sum = get_max_sum(arr2, 1, &amp;start, &amp;end); printf("sum: %d, start: %d, end: %d\n", sum, start, end); sum = get_max_sum(arr3, 4, &amp;start, &amp;end); printf("sum: %d, start: %d, end: %d\n", sum, start, end); sum = get_max_sum(arr4, 6, &amp;start, &amp;end); printf("sum: %d, start: %d, end: %d\n", sum, start, end); return 0; } $ ./a.out sum: 10, start: 3, end: 4 sum: 1, start: 0, end: 0 sum: -1, start: 0, end: 0 sum: 6, start: 3, end: 5 </code></pre> <p><strong>Update1</strong>: Added code to print the index of the subarray.</p> <p><strong>Update2</strong>: If two sub arrays with the same sum are found, choose the one with more elements.</p> <p><strong>Update3</strong>: Fix the algorithm for leading negative numbers</p>
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    1. POGiven an array of integers. Find the LARGEST subarray with the MAXIMUM sum
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    1. COI think this answer is correct as per my algorithm book.
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    2. COBut you are not giving the largest subarray, it is giving only the SUM.
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    3. COCode for getting the subarray is added. I thought it was not difficult so I didn't add it in the first place.
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