Given three integers N, A and B, the task is to find whether N can be represented as sum of A’s and B’s.
Examples:
Input: N = 11, A = 2, B = 3
Output: Yes
2 + 2 + 2 + 2 + 3 = 11Input: N = 8, A = 3, B = 7
Output: No
Approach: An efficient solution is to call a recursive function starting with zero (because zero is always possible). If function call is fun(x) then recursively call fun(x + a) and fun(x + b) (because if x is possible then x + a and x + b are also possible). Return out of the function if x > n.
Below is the implementation of the above approach:
C++
// CPP program to find if number N can// be represented as sum of a's and b's#include <bits/stdc++.h>using namespace std;// Function to find if number N can// be represented as sum of a's and b'svoid checkIfPossibleRec(int x, int a, int b, bool isPossible[], int n){ // base condition if (x > n) return; // if x is already visited if (isPossible[x]) return; // set x as possible isPossible[x] = true; // recursive call checkIfPossibleRec(x + a, a, b, isPossible, n); checkIfPossibleRec(x + b, a, b, isPossible, n);}bool checkPossible(int n, int a, int b){ bool isPossible[n + 1] = { false }; checkIfPossibleRec(0, a, b, isPossible, n); return isPossible[n];}// Driver programint main(){ int a = 3, b = 7, n = 8; if (checkPossible(a, b, n)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java program to find if number N can// be represented as sum of a's and b'simport java.util.*;class solution{// Function to find if number N can// be represented as sum of a's and b'sstatic void checkIfPossibleRec(int x, int a, int b, boolean isPossible[], int n){ // base condition if (x > n) return; // if x is already visited if (isPossible[x]) return; // set x as possible isPossible[x] = true; // recursive call checkIfPossibleRec(x + a, a, b, isPossible, n); checkIfPossibleRec(x + b, a, b, isPossible, n);}static boolean checkPossible(int n, int a, int b){ boolean isPossible[]=new boolean[n + 1]; for(int i=0;i<=n;i++) isPossible[i]=false; checkIfPossibleRec(0, a, b, isPossible, n); return isPossible[n];}// Driver programpublic static void main(String args[]){ int a = 3, b = 7, n = 8; if (checkPossible(a, b, n)) System.out.print("Yes"); else System.out.print( "No");}}//contributed by Arnab Kundu |
Python3
# Python3 program to find if number N can # be represented as sum of a's and b's # Function to find if number N can # be represented as sum of a's and b's def checkIfPossibleRec(x, a, b, isPossible, n): # base condition if x > n: return # If x is already visited if isPossible[x]: return # Set x as possible isPossible[x] = True # Recursive call checkIfPossibleRec(x + a, a, b, isPossible, n) checkIfPossibleRec(x + b, a, b, isPossible, n) def checkPossible(n, a, b): isPossible = [False] * (n + 1) checkIfPossibleRec(0, a, b, isPossible, n) return isPossible[n] # Driver Codeif __name__ == "__main__": a, b, n = 3, 7, 8 if checkPossible(a, b, n): print("Yes") else: print("No") # This code is contributed by Rituraj Jain |
C#
// C# program to find if number N can // be represented as sum of a's and b's using System;class GFG{// Function to find if number N can // be represented as sum of a's and b's static void checkIfPossibleRec(int x, int a, int b, bool []isPossible, int n) { // base condition if (x > n) return; // if x is already visited if (isPossible[x]) return; // set x as possible isPossible[x] = true; // recursive call checkIfPossibleRec(x + a, a, b, isPossible, n); checkIfPossibleRec(x + b, a, b, isPossible, n); } static bool checkPossible(int n, int a, int b) { bool []isPossible = new bool[n + 1]; for(int i = 0; i <= n; i++) isPossible[i] = false; checkIfPossibleRec(0, a, b, isPossible, n); return isPossible[n]; } // Driver Code static public void Main (){ int a = 3, b = 7, n = 8; if (checkPossible(a, b, n)) Console.WriteLine("Yes"); else Console.WriteLine( "No"); }}// This code is contributed by Sach_Code |
PHP
<?php// PHP program to find if number N can// be represented as sum of a's and b's// Function to find if number N can// be represented as sum of a's and b'sfunction checkIfPossibleRec($x, $a, $b, $isPossible, $n){ // base condition if ($x > $n) return; // if x is already visited if ($isPossible == true) return; // set x as possible $isPossible[$x] = true; // recursive call checkIfPossibleRec($x + $a, $a, $b, $isPossible, $n); checkIfPossibleRec($x + $b, $a, $b, $isPossible, $n);}function checkPossible($n, $a, $b){ $isPossible[$n + 1] = array(false); checkIfPossibleRec(0, $a, $b, $isPossible, $n); return $isPossible;}// Driver Code$a = 3;$b = 7;$n = 8;if (checkPossible($a, $b, $n)) echo "No";else echo "Yes";// This code is contributed by Sach_Code?> |
Javascript
<script>// Javascript program to find if number N can// be represented as sum of a's and b's // Function to find if number N can// be represented as sum of a's and b'sfunction checkIfPossibleRec(x, a, b, isPossible, n){ // Base condition if (x > n) return; // If x is already visited if (isPossible[x]) return; // Set x as possible isPossible[x] = true; // Recursive call checkIfPossibleRec(x + a, a, b, isPossible, n); checkIfPossibleRec(x + b, a, b, isPossible, n);}function checkPossible(n, a, b) { var isPossible = Array(n + 1).fill(false); checkIfPossibleRec(0, a, b, isPossible, n); return isPossible[n];}// Driver codevar a = 3, b = 7, n = 8;if (checkPossible(a, b, n)) document.write("Yes");else document.write("No");// This code is contributed by todaysgaurav </script> |
Output:
No
Time Complexity: O(2^n) , recursive function time complexity
Auxiliary Space: O(n), as extra space of size (n+1) is used to make a boolean array
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