Minimum removal of elements from end of an array required to obtain sum K

Given an integer K and an array A[] of size N, the task is to create a new array with sum K with minimum number of operations, where in each operation, an element can be removed either from the start or end of A[] and appended to the new array. If it is not possible to generate a new array with sum K, print -1. If there are multiple answers, print any one of them.

Examples

Input: K = 6, A[] = {1, 2, 3, 1, 3}, N = 5
Output: 1 3 2
Explanation: Operation 1: Removing A[0] modifies A[] to {2, 3, 1, 3}. Sum = 1.
Operation 2: Removing A[3] modifies A[] to {2, 1, 3}. Sum = 4.
Operation 3: Removing A[0] modifies A[] to {1, 3}. Sum = 6.

Input: K = 5, A[] = {1, 2, 7}, N = 3
Output: -1

Naive Approach: Follow the steps below to solve the problem:

1. The task is to find two minimum length subarrays, one from the beginning and one from the end of the array (possibly empty), such that their sum is equal to K.
2. Traverse the array from the left and calculate the subarray needed to be removed from the right such that the total sum is K.

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 minimum number of
// elements required to be removed from
// the ends of an array to obtain a sum K
int minSizeArr(int A[], int N, int K)
{
    // Number of elements removed from the
    // left and right ends of the array
    int leftTaken = N, rightTaken = N;

    // Sum of left and right subarrays
    int leftSum = 0, rightSum = 0;

    // No element is taken from left initially
    for (int left = -1; left < N; left++) {

        if (left != -1)
            leftSum += A[left];

        rightSum = 0;

        // Start taking elements from right side
        for (int right = N - 1; right > left; right--) {

            rightSum += A[right];

            if (leftSum + rightSum == K) {

                // (left + 1): Count of elements
                // removed from the left
                // (N-right): Count of elements
                // removed from the right
                if (leftTaken + rightTaken
                    > (left + 1) + (N - right)) {

                    leftTaken = left + 1;
                    rightTaken = N - right;
                }
                break;
            }
            // If sum is greater than K
            if (leftSum + rightSum > K)
                break;
        }
    }

    if (leftTaken + rightTaken <= N) {

        for (int i = 0; i < leftTaken; i++)
            cout << A[i] << " ";
        for (int i = 0; i < rightTaken; i++)
            cout << A[N - i - 1] << " ";
    }

    // If it is not possible to obtain sum K
    else
        cout << -1;
}

// Driver Code
int main()
{
    int N = 7;

    // Given Array
    int A[] = { 3, 2, 1, 1, 1, 1, 3 };

    // Given target sum
    int K = 10;

    minSizeArr(A, N, K);

    return 0;
}

Java

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


// Function to find the minimum number of
// elements required to be removed from
// the ends of an array to obtain a sum K
static void minSizeArr(int A[], int N, int K)
{
    // Number of elements removed from the
    // left and right ends of the array
    int leftTaken = N, rightTaken = N;

    // Sum of left and right subarrays
    int leftSum = 0, rightSum = 0;

    // No element is taken from left initially
    for (int left = -1; left < N; left++) {

        if (left != -1)
            leftSum += A[left];

        rightSum = 0;

        // Start taking elements from right side
        for (int right = N - 1; right > left; right--)
        {

            rightSum += A[right];

            if (leftSum + rightSum == K) {

                // (left + 1): Count of elements
                // removed from the left
                // (N-right): Count of elements
                // removed from the right
                if (leftTaken + rightTaken
                    > (left + 1) + (N - right)) {

                    leftTaken = left + 1;
                    rightTaken = N - right;
                }
                break;
            }
            // If sum is greater than K
            if (leftSum + rightSum > K)
                break;
        }
    }

    if (leftTaken + rightTaken <= N) {

        for (int i = 0; i < leftTaken; i++)
            System.out.print( A[i] + " ");
        for (int i = 0; i < rightTaken; i++)
            System.out.print(A[N - i - 1] + " ");
    }

    // If it is not possible to obtain sum K
    else
        System.out.print(-1);
}


// Driver code
public static void main(String[] args)
{
    int N = 7;

    // Given Array
    int A[] = { 3, 2, 1, 1, 1, 1, 3 };

    // Given target sum
    int K = 10;

    minSizeArr(A, N, K);
}
}

// This code is contributed by splevel62.

Python3

# Python3 program for the above approach

# Function to find the minimum number of
# elements required to be removed from
# the ends of an array to obtain a sum K
def minSizeArr(A, N, K):

    # Number of elements removed from the
    # left and right ends of the array
    leftTaken = N
    rightTaken = N

    # Sum of left and right subarrays
    leftSum = 0
    rightSum = 0

    # No element is taken from left initially
    for left in range(-1, N):
        if (left != -1):
            leftSum += A[left]

        rightSum = 0

        # Start taking elements from right side
        for right in range(N - 1, left, -1):
            rightSum += A[right]

            if (leftSum + rightSum == K):

                # (left + 1): Count of elements
                # removed from the left
                # (N-right): Count of elements
                # removed from the right
                if (leftTaken + rightTaken >
                   (left + 1) + (N - right)):
                    leftTaken = left + 1
                    rightTaken = N - right

                break

            # If sum is greater than K
            if (leftSum + rightSum > K):
                break

    if (leftTaken + rightTaken <= N):
        for i in range(leftTaken):
            print(A[i], end = " ")

        for i in range(rightTaken):
            print(A[N - i - 1], end = " ")

    # If it is not possible to obtain sum K
    else:
        print(-1)

# Driver Code
if __name__ == "__main__":

    N = 7

    # Given Array
    A = [ 3, 2, 1, 1, 1, 1, 3 ]

    # Given target sum
    K = 10

    minSizeArr(A, N, K)

# This code is contributed by ukasp

C#

// C# program for the above approach
using System;

class GFG {

// Function to find the smallest
// array that can be removed from
// the ends of an array to obtain sum K
static void minSizeArr(int[] A, int N, int K)
{

    // Number of elements removed from the
    // left and right ends of the array
    int leftTaken = N, rightTaken = N;

    // Sum of left and right subarrays
    int leftSum = 0, rightSum = 0;

    // No element is taken from left initially
    for (int left = -1; left < N; left++) {

        if (left != -1)
            leftSum += A[left];

        rightSum = 0;

        // Start taking elements from right side
        for (int right = N - 1; right > left; right--)
        {

            rightSum += A[right];

            if (leftSum + rightSum == K) {

                // (left + 1): Count of elements
                // removed from the left
                // (N-right): Count of elements
                // removed from the right
                if (leftTaken + rightTaken
                    > (left + 1) + (N - right)) {

                    leftTaken = left + 1;
                    rightTaken = N - right;
                }
                break;
            }
            // If sum is greater than K
            if (leftSum + rightSum > K)
                break;
        }
    }

    if (leftTaken + rightTaken <= N) {

        for (int i = 0; i < leftTaken; i++)
            Console.Write( A[i] + " ");
        for (int i = 0; i < rightTaken; i++)
            Console.Write(A[N - i - 1] + " ");
    }

    // If it is not possible to obtain sum K
    else
        Console.Write(-1);
}


    // Driver Code
    public static void Main()
    {
        int N = 7;

    // Given Array
    int[] A = { 3, 2, 1, 1, 1, 1, 3 };

    // Given target sum
    int K = 10;

    minSizeArr(A, N, K);
    }
}

// This code is contributed by code_hunt.

Javascript

<script>

// JavaScript program for the above approach

// Function to find the minimum number of
// elements required to be removed from
// the ends of an array to obtain a sum K
function minSizeArr(A, N, K)
{
    // Number of elements removed from the
    // left and right ends of the array
    let leftTaken = N, rightTaken = N;

    // Sum of left and right subarrays
    let leftSum = 0, rightSum = 0;

    // No element is taken from left initially
    for (let left = -1; left < N; left++) {

        if (left != -1)
            leftSum += A[left];

        rightSum = 0;

        // Start taking elements from right side
        for (let right = N - 1; right > left; right--)
        {

            rightSum += A[right];

            if (leftSum + rightSum == K) {

                // (left + 1): Count of elements
                // removed from the left
                // (N-right): Count of elements
                // removed from the right
                if (leftTaken + rightTaken
                    > (left + 1) + (N - right)) {

                    leftTaken = left + 1;
                    rightTaken = N - right;
                }
                break;
            }
            // If sum is greater than K
            if (leftSum + rightSum > K)
                break;
        }
    }

    if (leftTaken + rightTaken <= N) {

        for (let i = 0; i < leftTaken; i++)
            document.write( A[i] + " ");
        for (let i = 0; i < rightTaken; i++)
            document.write(A[N - i - 1] + " ");
    }

    // If it is not possible to obtain sum K
    else
        document.write(-1);
}

// Driver code
    let N = 7;

    // Given Array
    let A = [ 3, 2, 1, 1, 1, 1, 3 ];

    // Given target sum
    let K = 10;

    minSizeArr(A, N, K);

// This code is contributed by souraavghosh0416.
</script>

Output:

3 2 3 1 1

Time Complexity: O(N²)
Auxiliary Space: O(1)

Efficient Approach: Follow the steps below to optimize the above approach: 

1. Calculate the sum of elements of the array A[] and store it in a variable, say Total.
2. The problem can be seen as finding the maximum size subarray with sum (Total – K).
3. The remaining elements will add up to K.

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 smallest
// array that can be removed from
// the ends of an array to obtain sum K
void minSizeArr(int A[], int N, int K)
{
    int sum = 0;
    // Sum of complete array
    for (int i = 0; i < N; i++)
        sum += A[i];

    // If given number is greater
    // than sum of the array
    if (K > sum) {
        cout << -1;
        return;
    }

    // If number is equal to
    // the sum of array
    if (K == sum) {
        for (int i = 0; i < N; i++) {
            cout << A[i] << " ";
        }
        return;
    }

    // tar is sum of middle subarray
    int tar = sum - K;

    // Find the longest subarray
    // with sum equal to tar
    unordered_map<int, int> um;
    um[0] = -1;

    int left, right;
    int cur = 0, maxi = -1;
    for (int i = 0; i < N; i++) {
        cur += A[i];
        if (um.find(cur - tar) != um.end()
            && i - um[cur - tar] > maxi) {
            maxi = i - um[cur - tar];
            right = i;
            left = um[cur - tar];
        }
        if (um.find(cur) == um.end())
            um[cur] = i;
    }

    // If there is no subarray with
    // sum equal to tar
    if (maxi == -1)
        cout << -1;

    else {
        for (int i = 0; i <= left; i++)
            cout << A[i] << " ";
        for (int i = 0; i < right; i++)
            cout << A[N - i - 1] << " ";
    }
}

// Driver Code
int main()
{
    int N = 7;

    // Given Array
    int A[] = { 3, 2, 1, 1, 1, 1, 3 };

    // Given target sum
    int K = 10;

    minSizeArr(A, N, K);
    return 0;
}

Java

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

class GFG {

      // Function to find the smallest
// array that can be removed from
// the ends of an array to obtain sum K
static void minSizeArr(int A[], int N, int K)
{
    int sum = 0;

    // Sum of complete array
    for (int i = 0; i < N; i++)
        sum += A[i];

    // If given number is greater
    // than sum of the array
    if (K > sum) {
        System.out.print(-1);
        return;
    }

    // If number is equal to
    // the sum of array
    if (K == sum) {
        for (int i = 0; i < N; i++) {
            System.out.print(A[i] + " ");
        }
        return;
    }

    // tar is sum of middle subarray
    int tar = sum - K;

    // Find the longest subarray
    // with sum equal to tar
    HashMap<Integer, Integer> um = new HashMap<Integer, Integer>();
     um.put(0, -1);

    int left = 0, right = 0;
    int cur = 0, maxi = -1;
    for (int i = 0; i < N; i++) {
        cur += A[i];
        if (um.containsKey(cur - tar)
            && i - um.get(cur - tar) > maxi) {
            maxi = i - um.get(cur - tar);
            right = i;
            left = um.get(cur - tar);
        }
        if (!um.containsKey(cur))
            um.put(cur, i);
    }

    // If there is no subarray with
    // sum equal to tar
    if (maxi == -1)
        System.out.println(-1);

    else {
        for (int i = 0; i <= left; i++)
            System.out.print(A[i] + " ");
        for (int i = 0; i < right; i++)
            System.out.print(A[N - i - 1] + " ");
    }
}

// Driver Code
    public static void main (String[] args) {
        int N = 7;

    // Given Array
    int A[] = { 3, 2, 1, 1, 1, 1, 3 };

    // Given target sum
    int K = 10;

    minSizeArr(A, N, K);
    }
}

// This code is contributed by Dharanendra L V.

Python3

# python 3 program for the above approach

# Function to find the smallest
# array that can be removed from
# the ends of an array to obtain sum K
def minSizeArr(A, N, K):
    sum = 0

    # Sum of complete array
    for i in range(N):
        sum += A[i]

    # If given number is greater
    # than sum of the array
    if (K > sum):
        print(-1);
        return

    # If number is equal to
    # the sum of array
    if (K == sum):
        for i in range(N):
            print(A[i],end = " ")
        return

    # tar is sum of middle subarray
    tar = sum - K

    # Find the longest subarray
    # with sum equal to tar
    um = {}
    um[0] = -1

    left = 0
    right = 0
    cur = 0
    maxi = -1
    for i in range(N):
        cur += A[i]
        if((cur - tar) in um and (i - um[cur - tar]) > maxi):
            maxi = i - um[cur - tar]
            right = i
            left = um[cur - tar]
        if(cur not in um):
            um[cur] = i

    # If there is no subarray with
    # sum equal to tar
    if (maxi == -1):
        print(-1)

    else:
        for i in range(left+1):
            print(A[i], end = " ")
        for i in range(right):
            print(A[N - i - 1], end = " ")

# Driver Code
if __name__ == '__main__':
    N = 7

    # Given Array
    A = [3, 2, 1, 1, 1, 1, 3]

    # Given target sum
    K = 10
    minSizeArr(A, N, K)

    # This code is contributed by SURENDRA_GANGWAR.

C#

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

class GFG{

// Function to find the smallest
// array that can be removed from
// the ends of an array to obtain sum K
static void minSizeArr(int[] A, int N, int K)
{
    int sum = 0;

    // Sum of complete array
    for(int i = 0; i < N; i++)
        sum += A[i];

    // If given number is greater
    // than sum of the array
    if (K > sum)
    {
        Console.WriteLine(-1);
        return;
    }

    // If number is equal to
    // the sum of array
    if (K == sum)
    {
        for(int i = 0; i < N; i++)
        {
           Console.Write(A[i] + " ");
        }
        return;
    }

    // tar is sum of middle subarray
    int tar = sum - K;

    // Find the longest subarray
    // with sum equal to tar
    Dictionary<int,
               int> um =  new Dictionary<int,
                                         int>();
     um[0] = -1;

    int left = 0, right = 0;
    int cur = 0, maxi = -1;
    for(int i = 0; i < N; i++)
    {
        cur += A[i];
        if (um.ContainsKey(cur - tar) &&
             i - um[cur - tar] > maxi)
        {
            maxi = i - um[cur - tar];
            right = i;
            left = um[cur - tar];
        }
        if (!um.ContainsKey(cur))
            um[cur] = i;
    }

    // If there is no subarray with
    // sum equal to tar
    if (maxi == -1)
        Console.Write(-1);

    else
    {
        for(int i = 0; i <= left; i++)
            Console.Write(A[i] + " ");
        for(int i = 0; i < right; i++)
            Console.Write(A[N - i - 1] + " ");
    }
}

// Driver code
static public void Main()
{
    int N = 7;

    // Given Array
    int[] A = { 3, 2, 1, 1, 1, 1, 3 };

    // Given target sum
    int K = 10;

    minSizeArr(A, N, K);
}
}

// This code is contributed by offbeat

Javascript

<script>

// JavaScript program for the above approach

// Function to find the smallest
// array that can be removed from
// the ends of an array to obtain sum K
function minSizeArr(A, N, K)
{
    var sum = 0;
    var i;
    // Sum of complete array
    for (i = 0; i < N; i++)
        sum += A[i];

    // If given number is greater
    // than sum of the array
    if (K > sum) {
        cout << -1;
        return;
    }

    // If number is equal to
    // the sum of array
    if (K == sum) {
        for (i = 0; i < N; i++) {
            document.write(A[i]+' ');
        }
        return;
    }

    // tar is sum of middle subarray
    var tar = sum - K;

    // Find the longest subarray
    // with sum equal to tar
    var um = new Map();
    um[0] = -1;

    var left, right;
    var cur = 0, maxi = -1;
    for (i = 0; i < N; i++) {
        cur += A[i];
        if (um.has(cur - tar)
            && i - um.get(cur - tar) > maxi) {
            maxi = i - um.get(cur - tar);
            right = i;
            left = um.get(cur - tar);
        }
        if (!um.has(cur))
            um.set(cur,i);
    }

    // If there is no subarray with
    // sum equal to tar
    if (maxi == -1)
        cout << -1;

    else {
        for (i = 0; i <= left; i++)
            document.write(A[i]+' ');
        for (i = 0; i < right; i++)
            document.write(A[N - i - 1]+ ' ');
    }
}

// Driver Code

    var N = 7;

    // Given Array
    var A = [3, 2, 1, 1, 1, 1, 3];

    // Given target sum
    var K = 10;

    minSizeArr(A, N, K);

</script>

Output:

3 2 3 1 1

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

Related Topic: Subarrays, Subsequences, and Subsets in Array