Print the Maximum Subarray Sum

Given an array arr[], the task is to find the elements of a contiguous subarray of numbers which has the largest sum.

Examples:

Input: arr = [-2, -3, 4, -1, -2, 1, 5, -3]
Output: [4, -1, -2, 1, 5]
Explanation: 
In the above input the maximum contiguous subarray sum is 7 and the elements of the subarray are [4, -1, -2, 1, 5]

Input: arr = [-2, -5, 6, -2, -3, 1, 5, -6] 
Output: [6, -2, -3, 1, 5] 
Explanation: 
In the above input the maximum contiguous subarray sum is 7 and the elements 
of the subarray are [6, -2, -3, 1, 5]

Naive Approach: The naive approach is to generate all the possible subarray and print that subarray which has maximum sum. 
Time complexity: O(N2
Auxiliary Space: O(1)
Efficient Approach: The idea is to use the Kadane’s Algorithm to find the maximum subarray sum and store the starting and ending index of the subarray having maximum sum and print the subarray from starting index to ending index. Below are the steps:
 



  1. Initialize 3 variables endIndex to 0, currMax and globalMax to first value of the input array.
  2. For each element in the array starting from index(say i) 1, update currMax to max(nums[i], nums[i] + currMax) and globalMax and endIndex to i only if currMax > globalMax.
  3. To find the start index, iterate from endIndex in the left direction and keep decrementing the value of globalMax until it becomes 0. The point at which it becomes 0 is the start index.
  4. Now print the subarray between [start, end].

Below is the implementation of the above approach:
 

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the elements
// of Subarray with maximum sum
void SubarrayWithMaxSum(vector<int>& nums)
{
    // Initialize currMax and globalMax
    // with first value of nums
    int endIndex, currMax = nums[0];
    int globalMax = nums[0];
 
    // Iterate for all the elemensts
    // of the array
    for (int i = 1; i < nums.size(); ++i) {
 
        // Update currMax
        currMax = max(nums[i],
                      nums[i] + currMax);
 
        // Check if currMax is greater
        // than globalMax
        if (currMax > globalMax) {
            globalMax = currMax;
            endIndex = i;
        }
    }
 
    int startIndex = endIndex;
 
    // Traverse in left direction to
    // find start Index of subarray
    while (startIndex >= 0) {
 
        globalMax -= nums[startIndex];
 
        if (globalMax == 0)
            break;
 
        // Decrement the start index
        startIndex--;
    }
 
    // Printing the elements of
    // subarray with max sum
    for (int i = startIndex;
         i <= endIndex; ++i) {
 
        cout << nums[i] << " ";
    }
}
 
// Driver Code
int main()
{
    // Given array arr[]
    vector<int> arr
        = { -2, -5, 6, -2,
            -3, 1, 5, -6 };
 
    // Function call
    SubarrayWithMaxSum(arr);
    return 0;
}

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Java

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// Java program for the above approach
import java.util.*;
class GFG{
 
  // Function to print the elements
  // of Subarray with maximum sum
  static void SubarrayWithMaxSum(Vector<Integer> nums)
  {
    // Initialize currMax and globalMax
    // with first value of nums
    int endIndex = 0, currMax = nums.get(0);
    int globalMax = nums.get(0);
 
    // Iterate for all the elemensts
    // of the array
    for (int i = 1; i < nums.size(); ++i)
    {
 
      // Update currMax
      currMax = Math.max(nums.get(i),
                         nums.get(i) + currMax);
 
      // Check if currMax is greater
      // than globalMax
      if (currMax > globalMax)
      {
        globalMax = currMax;
        endIndex = i;
      }
    }
 
    int startIndex = endIndex;
 
    // Traverse in left direction to
    // find start Index of subarray
    while (startIndex >= 0)
    {
      globalMax -= nums.get(startIndex);
 
      if (globalMax == 0)
        break;
 
      // Decrement the start index
      startIndex--;
    }
 
    // Printing the elements of
    // subarray with max sum
    for(int i = startIndex; i <= endIndex; ++i)
    {
      System.out.print(nums.get(i) + " ");
    }
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    // Given array arr[]
    Vector<Integer> arr = new Vector<Integer>();
    arr.add(-2);
    arr.add(-5);
    arr.add(6);
    arr.add(-2);
    arr.add(-3);
    arr.add(1);
    arr.add(5);
    arr.add(-6);
 
    // Function call
    SubarrayWithMaxSum(arr);
  }
}
 
// This code is contributed by Rajput-Ji

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C#

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// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
 
  // Function to print the elements
  // of Subarray with maximum sum
  static void SubarrayWithMaxSum(List<int> nums)
  {
    // Initialize currMax and globalMax
    // with first value of nums
    int endIndex = 0, currMax = nums[0];
    int globalMax = nums[0];
 
    // Iterate for all the elemensts
    // of the array
    for (int i = 1; i < nums.Count; ++i)
    {
 
      // Update currMax
      currMax = Math.Max(nums[i],
                         nums[i] + currMax);
 
      // Check if currMax is greater
      // than globalMax
      if (currMax > globalMax)
      {
        globalMax = currMax;
        endIndex = i;
      }
    }
 
    int startIndex = endIndex;
 
    // Traverse in left direction to
    // find start Index of subarray
    while (startIndex >= 0)
    {
      globalMax -= nums[startIndex];
 
      if (globalMax == 0)
        break;
 
      // Decrement the start index
      startIndex--;
    }
 
    // Printing the elements of
    // subarray with max sum
    for(int i = startIndex; i <= endIndex; ++i)
    {
      Console.Write(nums[i] + " ");
    }
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    // Given array []arr
    List<int> arr = new List<int>();
    arr.Add(-2);
    arr.Add(-5);
    arr.Add(6);
    arr.Add(-2);
    arr.Add(-3);
    arr.Add(1);
    arr.Add(5);
    arr.Add(-6);
 
    // Function call
    SubarrayWithMaxSum(arr);
  }
}
 
// This code is contributed by gauravrajput1

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Output: 

6 -2 -3 1 5





 

Time complexity: O(N) 
Auxiliary Space: O(1)
 

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