Given an array of integers arr[] and two integers X and Y, the task is to check if it is possible to obtain a sequence having sum X such that the sum of each element of the subsequence multiplied by an array element is equal to Y.
Note: Here X is always less than Y.
Examples:
Input: arr[] = {1, 2, 7, 9, 10}, X = 11, Y = 13
Output: Yes
Explanation:
The given value of X( = 11) can be split into a sequence {9, 2} such that 9 * 1(= arr[0] + 2 * 2(= arr[1]) = 13( = Y)
Input: arr[] ={1, 3, 5, 7}, X = 27, Y = 34
Output: No
Approach: Follow the steps below in order to solve the problem:
- Calculate the difference between Y and X.
- For every array element arr[i], which is > 1, update (Y – X) % (arr[i] – 1).
- If the difference is reduced 0, print “Yes“. Otherwise, print “No”.
Below is the implementation of the above approach:
C++
// C++ Program to implement// the above approach#include <iostream>using namespace std;// Function to check if it is possible// to obtain sum Y from a sequence of// sum X from the array arr[]void solve(int arr[], int n, int X, int Y){ // Store the difference int diff = Y - X; // Iterate over the array for (int i = 0; i < n; i++) { if (arr[i] != 1) { diff = diff % (arr[i] - 1); } } // If diff reduced to 0 if (diff == 0) cout << "Yes"; else cout << "No";}// Driver Codeint main(){ int arr[] = { 1, 2, 7, 9, 10 }; int n = sizeof(arr) / sizeof(arr[0]); int X = 11, Y = 13; solve(arr, n, X, Y); return 0;} |
Java
// Java program to implement// the above approachclass GFG{ // Function to check if it is possible// to obtain sum Y from a sequence of// sum X from the array arr[]static void solve(int arr[], int n, int X, int Y){ // Store the difference int diff = Y - X; // Iterate over the array for(int i = 0; i < n; i++) { if (arr[i] != 1) { diff = diff % (arr[i] - 1); } } // If diff reduced to 0 if (diff == 0) System.out.print( "Yes"); else System.out.print("No");}// Driver Codepublic static void main (String []args){ int arr[] = { 1, 2, 7, 9, 10 }; int n = arr.length; int X = 11, Y = 13; solve(arr, n, X, Y);}}// This code is contributed by chitranayal |
Python3
# Python3 program to implement# the above approach# Function to check if it is possible# to obtain sum Y from a sequence of# sum X from the array arr[]def solve(arr, n, X, Y): # Store the difference diff = Y - X # Iterate over the array for i in range(n): if(arr[i] != 1): diff = diff % (arr[i] - 1) # If diff reduced to 0 if(diff == 0): print("Yes") else: print("No")# Driver Codearr = [ 1, 2, 7, 9, 10 ]n = len(arr)X, Y = 11, 13# Function callsolve(arr, n, X, Y)# This code is contributed by Shivam Singh |
C#
// C# program to implement// the above approachusing System;class GFG{ // Function to check if it is possible// to obtain sum Y from a sequence of// sum X from the array []arrstatic void solve(int []arr, int n, int X, int Y){ // Store the difference int diff = Y - X; // Iterate over the array for(int i = 0; i < n; i++) { if (arr[i] != 1) { diff = diff % (arr[i] - 1); } } // If diff reduced to 0 if (diff == 0) Console.Write("Yes"); else Console.Write("No");}// Driver Codepublic static void Main(String []args){ int []arr = { 1, 2, 7, 9, 10 }; int n = arr.Length; int X = 11, Y = 13; solve(arr, n, X, Y);}}// This code is contributed by Amit Katiyar |
Javascript
<script> // Javascript program to implement// the above approach// Function to check if it is possible// to obtain sum Y from a sequence of// sum X from the array arr[]function solve(arr, n, X, Y){ // Store the difference var diff = Y - X; // Iterate over the array for(var i = 0; i < n; i++) { if (arr[i] != 1) { diff = diff % (arr[i] - 1); } } // If diff reduced to 0 if (diff == 0) document.write("Yes"); else document.write("No");}// Driver Codevar arr = [ 1, 2, 7, 9, 10 ];var n = arr.length;var X = 11, Y = 13;solve(arr, n, X, Y);// This code is contributed by rutvik_56</script> |
Output:
Yes
Time Complexity: O(N)
Auxiliary Space: O(1)
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