{"post":"

Problem

Given an array of N integers. Find the contiguous sub-array with maximum sum.

Brute Force Approach

We will find the sum of all the subarrays and calculate the max, using two nested for loops.

#include<bits/stdc++.h>\nusing namespace std;\nvoid printLongestSumOfContinuousSubarray(int arr[], int arrSize){\nint sumGlobal = 0;\nfor(int i = 0; i < arrSize; i++){\nint sum = 0;\nfor(int j = i; j < arrSize; j++){\nsum += arr[j];\nsumGlobal = max(sumGlobal, sum);\n}\n}
\ncout&lt;&lt;\"Max Sum of contiguous subarray: \"&lt;&lt;sumGlobal;\n
\n}
\nint main(){\nint arr[] =  {-2, -3, 4, -1, -2, 1, 5, -3};\nint arrSize = sizeof(arr)/sizeof(arr[0]);\nprintLongestSumOfContinuousSubarray(arr, arrSize);\nreturn 0;\n}

TC – O(N²)

SC – O(1)

Optimal Approach

In optimized approach, we will use Kadane's algorithm, We'll keep a global maximum sum and a local maximum sum.

The local maximum sum at index i is the maximum of A[i] and the sum of A[i] and local maximum sum at index i-1.

#include<bits/stdc++.h>\nusing namespace std;\nvoid printLongestSumOfContinuousSubarray(int arr[], int arrSize){\nint localMax  = arr[0];\nint globalMax = arr[0];
\n  for(int i = 1; i &lt; arrSize; i++){\n    \tif(arr[i] &gt; localMax + arr[i]){\n\t\tlocalMax = arr[i];\n\t} else {\n\t\tlocalMax += arr[i];\n\t}\n      \n      \tif(localMax &gt; globalMax){\n\t\tglobalMax = localMax;\n\t}\n    }\n\ncout&lt;&lt;\"Max Sum of contiguous subarray: \"&lt;&lt;globalMax;\n
\n}
\nint main(){\nint arr[] =  {-2, -3, 4, -1, -2, 1, 5, -3};\nint arrSize = sizeof(arr)/sizeof(arr[0]);\nprintLongestSumOfContinuousSubarray(arr, arrSize);\nreturn 0;\n}

TC – O(N)

SC – O(1)

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