kadane's algorithm

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#include <bits/stdc++.h>

using namespace std;
int max_sumarray(int arr[],int n)//calculate the max sum; it has two parts
{
    int max_sum;
    int cur_sum;
    int count=0;
    for(int i=0;i<n;i++)
    {
        if(arr[i]<0)
        {
            count++;
        }
    }
    if(count!=n)//part 1 when elements are all +ve or +ve and -ve
    {
        max_sum=0;
        cur_sum=0;
        for(int i=0;i<n;i++)
        {
            cur_sum=cur_sum+arr[i];
            if(cur_sum>max_sum)
            {
                max_sum=cur_sum;
            }
            if(cur_sum<0)
            {
                cur_sum=0;
            }
        }
    }
    else if(count==n)//part 2 when all the elements are -ve
    {
        max_sum=arr[0];
        cur_sum=arr[0];
        for(int i=1;i<n;i++)
        {
            cur_sum=max(arr[i],cur_sum+arr[i]);
            max_sum=max(cur_sum,max_sum);
        }
    }
    return max_sum;
}

int main()
{
    int n;
    cout<<"enter the size of the array:"<<endl;
    cin>>n;
    int arr[n];
    cout<<"enter the elements of the array:"<<endl;
    for(int i=0;i<n;i++)
    {
        cin>>arr[i];
    }
    int max_subarray_sum=max_sumarray(arr,n);
    cout<<max_subarray_sum<<endl;
}

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Contributed on Jun 16 2021

ujjwal sotra

0 Answers Avg Quality 2/10

kadane's algorithm

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3

let array = [-3, 1, -8, 12, 0, -3, 5, -9, 4]

// We are using sliding window technique
// Time complexity: O(n), Space: O(1)
function maxSubArray(array) {
    let current_sum = array[0]
    let max_sum = array[0]

    for (let i = 1; i < array.length; i++) {
        current_sum = Math.max(array[i], current_sum + array[i])

        if (current_sum > max_sum) {
            max_sum = current_sum
        }
    }

    return max_sum
}

console.log(maxSubArray(array))		// [Log]: 14

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Contributed on Aug 18 2022

Mysterious Macaque

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kadane's algorithm code

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#include <bits/stdc++.h>
using namespace std;

int main()
{
    int a[]={2, -4, 3, -1, 2};
    int local_max = 0;
    int global_max = INT_MIN;
    int n=sizeof(a)/sizeof(a[0]);
    
    for(i = 0; i < n;i++)
	{   
	    local_max = max(a[i], a[i] + local_max);
        
	    if(local_max > global_max)
	    {
	        global_max = local_max; 
	    }
	    
	}
	cout<<global_max;
	return 0;
}

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Contributed on Aug 20 2022

Proud Pig

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kadane algorithm

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6

public int kadane(int[] arr){
	int max_so_far = 0, curr_max = Integer.MIN_VALUE;
    for(int i: arr){
    	max_so_far += i;
        if(max_so_far<0) max_so_far = 0;
        if(max_so_far>curr_max) curr_max = max_so_far;
    }
    return curr_max;
}

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Contributed on Jun 20 2020

Crowded Civet

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kadane's algorithm

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1

class Solution {
public:
    int maxProduct(vector<int>& nums) {
        // Return early if input is empty
        if (nums.empty()) return 0;
		
        int best = INT_MIN;
        // Running products
        int maxProd = 1;
        int minProd = 1;
        
        for (int n: nums) {
            if (n < 0) {
                // Swap max and min
                swap(maxProd, minProd);
            }
            // Reset to current value if smaller or larger than it
            // (intuitively means that we start considering a new sub-array)
            maxProd = max(maxProd*n, n);
            minProd = min(minProd*n, n);
            // Update the best
            best = max(best, maxProd);
        }
        
        return best;
    }
};

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Contributed on Jun 21 2022

Jolly Jellyfish

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kadane's algorithm

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public class Kadanes_Algorithm {
    public static void kadane(int[] Array) {
        int Max_Sum = Integer.MIN_VALUE;
        int Current_Sum = 0;

        for (int i = 0; i < Array.length; i++) {
            Current_Sum += Array[i];

            if (Current_Sum < 0) {
                Current_Sum = 0;
            }

            Max_Sum = Math.max(Max_Sum, Current_Sum);
        }
        System.out.println("Max Sum : " + Max_Sum);
    }

    public static void main(String[] args) {
        int[] arr = {-1, -2, 6, -1, 3};

        kadane(arr);
    }
}

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Contributed on Feb 03 2023

Poor Tony Stark

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kadane's Algorithem

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0

// C++ program to print largest contiguous array sum
#include<iostream>
#include<climits>
using namespace std;
  
int maxSubArraySum(int a[], int size)
{
    int max_so_far = INT_MIN, max_ending_here = 0;
  
    for (int i = 0; i < size; i++)
    {
        max_ending_here = max_ending_here + a[i];
        if (max_so_far < max_ending_here)
            max_so_far = max_ending_here;
  
        if (max_ending_here < 0)
            max_ending_here = 0;
    }
    return max_so_far;
}
  
/*Driver program to test maxSubArraySum*/
int main()
{
    int a[] = {-2, -3, 4, -1, -2, 1, 5, -3};
    int n = sizeof(a)/sizeof(a[0]);
    int max_sum = maxSubArraySum(a, n);
    cout << "Maximum contiguous sum is " << max_sum;
    return 0;
}

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Contributed on Aug 31 2022

MOhammed Bilaal Arham

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kadane algorithm

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0

Lets take the example:
    {-2, -3, 4, -1, -2, 1, 5, -3}

    max_so_far = max_ending_here = 0

    for i=0,  a[0] =  -2
    max_ending_here = max_ending_here + (-2)
    Set max_ending_here = 0 because max_ending_here < 0

    for i=1,  a[1] =  -3
    max_ending_here = max_ending_here + (-3)
    Set max_ending_here = 0 because max_ending_here < 0

    for i=2,  a[2] =  4
    max_ending_here = max_ending_here + (4)
    max_ending_here = 4
    max_so_far is updated to 4 because max_ending_here greater 
    than max_so_far which was 0 till now

    for i=3,  a[3] =  -1
    max_ending_here = max_ending_here + (-1)
    max_ending_here = 3

    for i=4,  a[4] =  -2
    max_ending_here = max_ending_here + (-2)
    max_ending_here = 1

    for i=5,  a[5] =  1
    max_ending_here = max_ending_here + (1)
    max_ending_here = 2

    for i=6,  a[6] =  5
    max_ending_here = max_ending_here + (5)
    max_ending_here = 7
    max_so_far is updated to 7 because max_ending_here is 
    greater than max_so_far

    for i=7,  a[7] =  -3
    max_ending_here = max_ending_here + (-3)
    max_ending_here = 4

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Contributed on Dec 30 2021

Magnificent Mink

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Kadane's Algorithm - I

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class KadaneAlgo1
{
    //Function to print the maximum contiguous subarray sum ending at each 
    //position in the array and return the overall maximum.
    public long maximumSum(int arr[], int sizeOfArray)
    {
        int res=arr[0];
        int currmax=arr[0];
        StringBuffer sb=new StringBuffer();
        sb.append(arr[0]+" ");
        for(int i=1;i<sizeOfArray;i++){
            currmax=Math.max(currmax+arr[i],arr[i]);
            res=Math.max(res,currmax);
            sb.append(currmax+" ");
        }
        System.out.println(sb.toString().trim());
        return res;
    }

}

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Contributed on Oct 19 2022

Sumit Rawal

0 Answers Avg Quality 2/10

Kadane's Algorithm - II

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class KadanesAlgorithm2
{
    //Function to return the maximum sum without adding adjacent elements.
    public long maximumSum(int arr[], int sizeOfArray)
    {
        if(sizeOfArray==1)
            return arr[0];
        if(sizeOfArray==2)
            return Math.max(arr[0],arr[1]);

        long[] dp=new long[sizeOfArray+1];
        dp[1]=arr[0];
        dp[2]=Math.max(arr[0],arr[1]);
        for(int i=3;i<=sizeOfArray;i++){
            dp[i]=Math.max(arr[i-1],Math.max(arr[i-1]+dp[i-2],dp[i-1]));
        }
        return dp[sizeOfArray];
    }

}

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Contributed on Oct 19 2022

Sumit Rawal

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kadane's algorithm

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