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:
- 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.
- 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> |
3 2 3 1 1
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: Follow the steps below to optimize the above approach:
- Calculate the sum of elements of the array A[] and store it in a variable, say Total.
- The problem can be seen as finding the maximum size subarray with sum (Total – K).
- 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> |
3 2 3 1 1
Time Complexity: O(N)
Auxiliary Space: O(N)
Related Topic: Subarrays, Subsequences, and Subsets in Array