Smallest sum contiguous subarray | Set-2

Given an array containing N integers. The task is to find the sum of the elements of the contiguous subarray having the smallest(minimum) sum.

Examples:

Input: arr[] = {3, -4, 2, -3, -1, 7, -5}
Output:-6

Input: arr = {2, 6, 8, 1, 4}
Output: 1

An approach has already been discussed in the previous post. In this post, a solution using the approach of Largest Sum Contiguous Subarray is discussed. This is based on the fact that in order to find the minimum contiguous sum we can first make the elements of the original array negative ie. Replace each ar[i] by -ar[i] and then apply Kadane Algorithm. Clearly, if this is the max sum formed then the minimum sum would be the negative of this sum.



Below is the implementation of above approach:

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// C++ program for
// Smallest sum contiguous subarray | Set 2
#include <bits/stdc++.h>
  
using namespace std;
  
// function to find the smallest sum contiguous subarray
int smallestSumSubarr(int arr[], int n)
{
    // First invert the sign of the elements
    for (int i = 0; i < n; i++)
        arr[i] = -arr[i];
  
    // Apply the normal Kadane algorithm But on the elements
    // Of the array having inverted sign
    int sum_here = arr[0], max_sum = arr[0];
  
    for (int i = 1; i < n; i++) {
  
        sum_here = max(sum_here + arr[i], arr[i]);
        max_sum = max(max_sum, sum_here);
    }
  
    // Invert  the answer to get minimum val
    return (-1) * max_sum;
}
  
// Driver Code
int main()
{
    int arr[] = { 3, -4, 2, -3, -1, 7, -5 };
  
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << "Smallest sum: "
         << smallestSumSubarr(arr, n);
    return 0;
}
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// Java program for Smallest
// sum contiguous subarray | Set 2
import java.io.*;
  
class GFG 
{
  
// function to find the 
// smallest sum contiguous
// subarray
static int smallestSumSubarr(int arr[],
                             int n)
{
    // First invert the 
    // sign of the elements
    for (int i = 0; i < n; i++)
        arr[i] = -arr[i];
  
    // Apply the normal Kadane 
    // algorithm But on the 
    // elements Of the array 
    // having inverted sign
    int sum_here = arr[0], 
        max_sum = arr[0];
  
    for (int i = 1; i < n; i++) 
    {
        sum_here = Math.max(sum_here + 
                            arr[i], arr[i]);
        max_sum = Math.max(max_sum, 
                           sum_here);
    }
  
    // Invert the answer
    // to get minimum val
    return (-1) * max_sum;
}
  
// Driver Code
public static void main (String[] args) 
{
    int arr[] = {3, -4, 2, -3
                -1, 7, -5};
  
    int n = arr.length;
  
    System.out.print("Smallest sum: "
            smallestSumSubarr(arr, n));
}
}
  
// This code is contributed 
// by inder_verma.
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# Python3 program for
# Smallest sum contiguous subarray | Set 2
  
# function to find the smallest 
# sum contiguous subarray
def smallestSumSubarr(arr, n):
      
    # First invert the sign of the elements
    for i in range(n):
        arr[i] = -arr[i]
  
    # Apply the normal Kadane algorithm but 
    # on the elements of the array having inverted sign
    sum_here = arr[0]
    max_sum = arr[0]
  
    for i in range(1, n):
  
        sum_here = max(sum_here + arr[i], arr[i])
        max_sum = max(max_sum, sum_here)
  
    # Invert the answer to get minimum val
    return (-1) * max_sum
  
# Driver Code
arr = [3, -4, 2, -3, -1, 7, -5]
  
n = len(arr)
  
print("Smallest sum:"
       smallestSumSubarr(arr, n))
  
# This code is contributed by Mohit Kumar
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// C# program for Smallest
// sum contiguous subarray | Set 2
using System;
class GFG 
{
  
// function to find the 
// smallest sum contiguous
// subarray
static int smallestSumSubarr(int []arr,
                             int n)
{
    // First invert the 
    // sign of the elements
    for (int i = 0; i < n; i++)
        arr[i] = -arr[i];
  
    // Apply the normal Kadane 
    // algorithm But on the 
    // elements Of the array 
    // having inverted sign
    int sum_here = arr[0], 
        max_sum = arr[0];
  
    for (int i = 1; i < n; i++) 
    {
        sum_here = Math.Max(sum_here + 
                            arr[i], arr[i]);
        max_sum = Math.Max(max_sum, 
                           sum_here);
    }
  
    // Invert the answer
    // to get minimum val
    return (-1) * max_sum;
}
  
// Driver Code
public static void Main () 
{
    int []arr = {3, -4, 2, -3, 
                -1, 7, -5};
  
    int n = arr.Length;
  
    Console.WriteLine("Smallest sum: "
             smallestSumSubarr(arr, n));
}
}
  
// This code is contributed 
// by inder_verma.
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<?php
// PHP program for Smallest sum
// contiguous subarray | Set 2 
  
// Function to find the smallest 
// sum contiguous subarray 
function smallestSumSubarr($arr, $n
    // First invert the sign 
    // of the elements 
    for ( $i = 0; $i < $n; $i++) 
        $arr[$i] = -$arr[$i]; 
  
    // Apply the normal Kadane algorithm
    // but on the elements of the array
    // having inverted sign 
    $sum_here = $arr[0];
    $max_sum = $arr[0]; 
  
    for ($i = 1; $i < $n; $i++) 
    
        $sum_here = max($sum_here
                        $arr[$i], $arr[$i]); 
        $max_sum = max($max_sum, $sum_here); 
    
  
    // Invert the answer to 
    // get minimum val 
    return (-1) * $max_sum
  
// Driver Code 
$arr = array( 3, -4, 2, -3, -1, 7, -5 ); 
  
$n = sizeof($arr); 
  
echo "Smallest sum: ",
    smallestSumSubarr($arr, $n); 
  
// This code is contributed 
// by Sach_Code
?>
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Output:
Smallest sum: -6

Time Complexity: O(n)




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