# Maximum sum subarray | Kadane’s Algo

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## 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>
using namespace std;

void printLongestSumOfContinuousSubarray(int arr[], int arrSize){
int sumGlobal = 0;
for(int i = 0; i < arrSize; i++){
int sum = 0;
for(int j = i; j < arrSize; j++){
sum += arr[j];
sumGlobal = max(sumGlobal, sum);
}
}

cout<<"Max Sum of contanuous subarray: "<<sumGlobal;
}

int main(){
int arr[] =  {-2, -3, 4, -1, -2, 1, 5, -3};
int arrSize = sizeof(arr)/sizeof(arr[0]);
printLongestSumOfContinuousSubarray(arr, arrSize);
return 0;
}``````

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>
using namespace std;

void printLongestSumOfContinuousSubarray(int arr[], int arrSize){
int localMax  = arr[0];
int globalMax = arr[0];

for(int i = 1; i < arrSize; i++){
if(arr[i] > localMax + arr[i]){
localMax = arr[i];
} else {
localMax += arr[i];
}

if(localMax > globalMax){
globalMax = localMax;
}
}

cout<<"Max Sum of contanuous subarray: "<<globalMax;
}

int main(){
int arr[] =  {-2, -3, 4, -1, -2, 1, 5, -3};
int arrSize = sizeof(arr)/sizeof(arr[0]);
printLongestSumOfContinuousSubarray(arr, arrSize);
return 0;
}``````

TC – O(N)

SC – O(1)

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