Maximum sum subarray | Kadane’s Algo

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&lt;&lt;"Max Sum of contanuous subarray: "&lt;&lt;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 &lt; arrSize; i++){
    	if(arr[i] &gt; localMax + arr[i]){
		localMax = arr[i];
	} else {
		localMax += arr[i];
	}
      
      	if(localMax &gt; globalMax){
		globalMax = localMax;
	}
    }

cout&lt;&lt;"Max Sum of contanuous subarray: "&lt;&lt;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)