Minimum array elements required to be subtracted from either end to reduce K to 0

Given an array arr[] consisting of N integers and an integer K, the task is to reduce K to 0 by removing an array element from either end of the array and subtracting it from K. If it is impossible to reduce K to 0, then print “-1”. Otherwise, print the minimum number of such operations required.

Examples:

Input: arr[] = {1, 3, 1, 1, 2}, K = 4
Output: 2
Explanation:
The given array is {1, 3, 1, 1, 2}
Operation1: Removing arr[0] (= 1) modifies arr[] to {3, 1, 1, 2}. Subtracting arr[0] from K reduces K to 4 – 1 = 3.
Operation2: Removing arr[0] (= 1) modifies arr[] to {1, 1, 2}. Subtracting arr[0] from K reduces K to 3 – 3 = 0.
Therefore, the total number of steps required is 2.

Input: arr[] = {1, 1, 3, 4}, K = 3
Output: -1

Approach: The given problem can be solved based on the following observations:

• Consider that X and Y elements are required to be removed from the front and back of the given array respectively, to reduce K to 0.
• Therefore, sum of the remaining array elements must be equal to (sum of array elements – K).

Therefore, from the above observation, the idea is to find the maximum length of subarray (say L) whose sum is equal to (sum of array elements – K). Therefore, print the value of (N – L) as the resultant minimum element that must be removed to reduce K to 0. If there doesn’t exist any such subarray, then print “-1”.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the length of
// longest subarray having sum K
int longestSubarray(int arr[],
                    int N, int K)
{
    // Stores the index of
    // the prefix sum
    unordered_map<int, int> um;

    int sum = 0, maxLen = 0;

    // Traverse the given array
    for (int i = 0; i < N; i++) {

        // Update sum
        sum += arr[i];

        // If the subarray starts
        // from index 0
        if (sum == K)
            maxLen = i + 1;

        // Add the current prefix sum
        // with index if it is not
        // present in the map um
        if (um.find(sum) == um.end())
            um[sum] = i;

        // Check if sum - K is
        // present in Map um or not
        if (um.find(sum - K) != um.end()) {

            // Update the maxLength
            if (maxLen < (i - um[sum - K]))
                maxLen = i - um[sum - K];
        }
    }

    // Return the required maximum length
    return maxLen;
}

// Function to find the minimum removal of
// array elements required to reduce K to 0
void minRequiredOperation(int arr[],
                          int N, int K)
{
    // Stores the sum of the array
    int TotalSum = 0;

    // Traverse the array arr[]
    for (int i = 0; i < N; i++)

        // Update sum of the array
        TotalSum += arr[i];

    // Find maxLen
    int maxLen = longestSubarray(
        arr, N, TotalSum - K);

    // If the subarray with
    // sum doesn't exist
    if (maxLen == -1) {
        cout << -1;
    }

    // Otherwise, print the
    // required maximum length
    else
        cout << N - maxLen;
}

// Driver Code
int main()
{
    int arr[] = { 1, 3, 1, 1, 2 };
    int K = 4;
    int N = sizeof(arr) / sizeof(arr[0]);

    minRequiredOperation(arr, N, K);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to find the length of
// longest subarray having sum K
static int longestSubarray(int[] arr, int N, int K)
{

    // Stores the index of
    // the prefix sum
    HashMap<Integer,
            Integer> um = new HashMap<Integer,
                                      Integer>();

    int sum = 0, maxLen = 0;

    // Traverse the given array
    for(int i = 0; i < N; i++)
    {

        // Update sum
        sum += arr[i];

        // If the subarray starts
        // from index 0
        if (sum == K)
            maxLen = i + 1;

        // Add the current prefix sum
        // with index if it is not
        // present in the map um
        if (!um.containsKey(sum))
            um.put(sum, i);

        // Check if sum - K is
        // present in Map um or not
        if (um.containsKey(sum - K))
        {

            // Update the maxLength
            if (maxLen < (i - um.get(sum - K)))
                maxLen = i - um.get(sum - K);
        }
    }

    // Return the required maximum length
    return maxLen;
}

// Function to find the minimum removal of
// array elements required to reduce K to 0
static void minRequiredOperation(int[] arr, int N,
                                 int K)
{

    // Stores the sum of the array
    int TotalSum = 0;

    // Traverse the array arr[]
    for(int i = 0; i < N; i++)

        // Update sum of the array
        TotalSum += arr[i];

    // Find maxLen
    int maxLen = longestSubarray(arr, N, TotalSum - K);

    // If the subarray with
    // sum doesn't exist
    if (maxLen == -1)
    {
        System.out.println(-1);
    }

    // Otherwise, print the
    // required maximum length
    else
        System.out.println(N - maxLen);
}

// Driver Code
public static void main(String[] args)
{
    int[] arr = { 1, 3, 1, 1, 2 };
    int K = 4;
    int N = arr.length;

    minRequiredOperation(arr, N, K);
}
}

// This code is contributed by ukasp

Python3

# Python3 program for the above approach

# Function to find the length of
# longest subarray having sum K
def longestSubarray(arr, N, K):

  # Stores the index of
    # the prefix sum
    um = {}
    sum , maxLen = 0, 0

    # Traverse the given array
    for i in range(N):

        # Update sum
        sum += arr[i]

        # If the subarray starts
        # from index 0
        if (sum == K):
            maxLen = i + 1

        # Add the current prefix sum
        # with index if it is not
        # present in the map um
        if (sum not in um):
            um[sum] = i

        # Check if sum - K is
        # present in Map um or not
        if ((sum - K) in um):

            # Update the maxLength
            if (maxLen < (i - um[sum - K])):
                maxLen = i - um[sum - K]

    # Return the required maximum length
    return maxLen

# Function to find the minimum removal of
# array elements required to reduce K to 0
def minRequiredOperation(arr, N, K):

    # Stores the sum of the array
    TotalSum = 0

    # Traverse the array arr[]
    for i in range(N):

      # Update sum of the array
        TotalSum += arr[i]

    # Find maxLen
    maxLen = longestSubarray(arr, N, TotalSum - K)

    # If the subarray with
    # sum doesn't exist
    if (maxLen == -1):
        print (-1,end="")

    # Otherwise, print the
    # required maximum length
    else:
        print (N - maxLen,end="")

# Driver Code
if __name__ == '__main__':
    arr = [1, 3, 1, 1, 2]
    K = 4
    N = len(arr)

    minRequiredOperation(arr, N, K)

    # this code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG
{

  // Function to find the length of
  // longest subarray having sum K
  static int longestSubarray(int []arr,
                             int N, int K)
  {
    // Stores the index of
    // the prefix sum
    Dictionary<int,int> um = new Dictionary<int,int>();

    int sum = 0, maxLen = 0;

    // Traverse the given array
    for (int i = 0; i < N; i++) {

      // Update sum
      sum += arr[i];

      // If the subarray starts
      // from index 0
      if (sum == K)
        maxLen = i + 1;

      // Add the current prefix sum
      // with index if it is not
      // present in the map um
      if (!um.ContainsKey(sum))
        um[sum] = i;

      // Check if sum - K is
      // present in Map um or not
      if (um.ContainsKey(sum - K)) {

        // Update the maxLength
        if (maxLen < (i - um[sum - K]))
          maxLen = i - um[sum - K];
      }
    }

    // Return the required maximum length
    return maxLen;
  }

  // Function to find the minimum removal of
  // array elements required to reduce K to 0
  static void minRequiredOperation(int []arr,
                                   int N, int K)
  {

    // Stores the sum of the array
    int TotalSum = 0;

    // Traverse the array arr[]
    for (int i = 0; i < N; i++)

      // Update sum of the array
      TotalSum += arr[i];

    // Find maxLen
    int maxLen = longestSubarray(
      arr, N, TotalSum - K);

    // If the subarray with
    // sum doesn't exist
    if (maxLen == -1) {
      Console.WriteLine(-1);
    }

    // Otherwise, print the
    // required maximum length
    else
      Console.WriteLine(N - maxLen);
  }

  // Driver Code
  public static void Main()
  {
    int []arr = { 1, 3, 1, 1, 2 };
    int K = 4;
    int N = arr.Length;
    minRequiredOperation(arr, N, K);
  }
}

// This code is contributed by SUREDRA_GANGWAR.

Javascript

<script>

// Javascript program for the above approach

// Function to find the length of
// longest subarray having sum K
function longestSubarray(arr, N, K)
{

    // Stores the index of
    // the prefix sum
    let um = new Map();

    let sum = 0, maxLen = 0;

    // Traverse the given array
    for(let i = 0; i < N; i++)
    {

        // Update sum
        sum += arr[i];

        // If the subarray starts
        // from index 0
        if (sum == K)
            maxLen = i + 1;

        // Add the current prefix sum
        // with index if it is not
        // present in the map um
        if (!um.has(sum))
            um.set(sum, i);

        // Check if sum - K is
        // present in Map um or not
        if (um.has(sum - K))
        {

            // Update the maxLength
            if (maxLen < (i - um.get(sum - K)))
                maxLen = i - um.get(sum - K);
        }
    }

    // Return the required maximum length
    return maxLen;
}

// Function to find the minimum removal of
// array elements required to reduce K to 0
function minRequiredOperation(arr, N, K)
{

    // Stores the sum of the array
    let TotalSum = 0;

    // Traverse the array arr[]
    for(let i = 0; i < N; i++)

        // Update sum of the array
        TotalSum += arr[i];

    // Find maxLen
    let maxLen = longestSubarray(
            arr, N, TotalSum - K);

    // If the subarray with
    // sum doesn't exist
    if (maxLen == -1)
    {
        document.write(-1);
    }

    // Otherwise, print the
    // required maximum length
    else
        document.write(N - maxLen);
}

// Driver Code
let arr = [ 1, 3, 1, 1, 2 ];
let K = 4;
let N = arr.length

minRequiredOperation(arr, N, K);

// This code is contributed by gfgking

</script>

Output:

2

Time Complexity: O(N)
Auxiliary Space: O(N)