Given an integer N, the task is to find two numbers a and b such that a + b = N, where a and b doesn’t contain any of the digits as K. Print -1 if not possible.
Examples:
Input: N = 100, K = 0
Output: 1 99
Explanation:
1 + 99 = 100 and none of the numbers has 0 in it.Input: N = 123456789, K = 2
Output: 3456790 119999999
Explanation:
3456790 + 119999999 = 123456789 and none of the numbers has 2 in it.
Approach: The idea is to iterate a loop from i = 1 to n-1, and check if i and (n – i) do not contain K. If it doesn’t contain any digit as then print the two numbers and break out of the loop.
For Example: N = 11, k = 0
i = 1, N – 1 = 10 but 10 contains 0, so increment i
i = 2, N – 1 = 9, so print this i and n – i.
Below is the implementation of the above approach:
C++
// C++ program for // the above approach#include<bits/stdc++.h>using namespace std;int freqCount(string str, char k){ int count = 0; for(int i = 0; i < str.size(); i++) { if (str[i] == k) count++; } return count;}// Function to find two // numbers whose sum// is N and do not // contain any digit as kvoid findAandB(int n, int k){ int flag = 0; // Check every number i and (n-i) for(int i = 1; i < n; i++) { // Check if i and n-i doesn't // contain k in them print i and n-i if (freqCount(to_string(i), (char)(k + 48)) == 0 and freqCount(to_string(n - i), (char)(k + 48)) == 0) { cout << "(" << i << ", " << n - i << ")"; flag = 1; break; } } // Check if flag is 0 // then print -1 if (flag == 0) cout << -1;}// Driver Codeint main(){ // Given N and K int N = 100; int K = 0; // Function call findAandB(N, K); return 0;}// This code is contributed by Rajput-Ji |
Java
// Java program for the above approachimport java.util.*;class GFG{static int freqCount(String str, char k){ int count = 0; for(int i = 0; i < str.length(); i++) { if (str.charAt(i) == k) count++; } return count;}// Function to find two numbers // whose sum is N and do not// contain any digit as kstatic void findAandB(int n, int k){ int flag = 0; // Check every number i and (n-i) for(int i = 1; i < n; i++) { // Check if i and n-i doesn't // contain k in them print i and n-i if (freqCount(Integer.toString(i), (char)(k + 48)) == 0 && freqCount(Integer.toString(n - i), (char)(k + 48)) == 0) { System.out.print("(" + i + ", " + (n - i) + ")"); flag = 1; break; } } // Check if flag is 0 // then print -1 if (flag == 0) System.out.print(-1);}// Driver codepublic static void main(String[] args){ // Given N and K int N = 100; int K = 0; // Function call findAandB(N, K);}}// This code is contributed by offbeat |
Python
# Python program for the above approach# Function to find two numbers whose sum# is N and do not contain any digit as kdef findAandB(n, k): flag = 0 # Check every number i and (n-i) for i in range(1, n): # Check if i and n-i doesn't # contain k in them print i and n-i if str(i).count(chr(k + 48)) == 0 \ and str(n-i).count(chr(k + 48)) == 0: print(i, n-i) flag = 1 break # check if flag is 0 then print -1 if(flag == 0): print(-1)# Driver Codeif __name__ == '__main__': # Given N and K N = 100 K = 0 # Function Call findAandB(N, K) |
C#
// C# program for the // above approachusing System;class GFG{static int freqCount(String str, char k){ int count = 0; for(int i = 0; i < str.Length; i++) { if (str[i] == k) count++; } return count;}// Function to find two numbers // whose sum is N and do not// contain any digit as kstatic void findAandB(int n, int k){ int flag = 0; // Check every number i and (n-i) for(int i = 1; i < n; i++) { // Check if i and n-i doesn't // contain k in them print i and n-i if (freqCount(i.ToString(), (char)(k + 48)) == 0 && freqCount((n-i).ToString(), (char)(k + 48)) == 0) { Console.Write("(" + i + ", " + (n - i) + ")"); flag = 1; break; } } // Check if flag is 0 // then print -1 if (flag == 0) Console.Write(-1);}// Driver codepublic static void Main(String[] args){ // Given N and K int N = 100; int K = 0; // Function call findAandB(N, K);}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program for // the above approachfunction freqCount(str, k){ var count = 0; for(var i = 0; i < str.length; i++) { if (str[i] == k) count++; } return count;}// Function to find two // numbers whose sum// is N and do not // contain any digit as kfunction findAandB(n, k){ var flag = 0; // Check every number i and (n-i) for(var i = 1; i < n; i++) { // Check if i and n-i doesn't // contain k in them print i and n-i if (freqCount(i.toString(), String.fromCharCode(k + 48)) == 0 && freqCount((n - i).toString(), String.fromCharCode(k + 48)) == 0) { document.write( "(" + i + ", " + (n - i) + ")"); flag = 1; break; } } // Check if flag is 0 // then print -1 if (flag == 0) cout + -1;}// Driver Code// Given N and Kvar N = 100;var K = 0;// Function callfindAandB(N, K);// This code is contributed by rrrtnx.</script> |
Output:
(1, 99)
Time Complexity: O(N*L) where L is the length of N
Auxiliary Space: O(1)
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