Given a number 
and two digits 
and 
. The task is to find the least number not less than N which contains the equal number of digits A and B.
Note: N <= 107
Examples:
Input : N = 4500, A = 4, B = 7
Output : 4747
The number greater than 4500 which has the same quantity of number ‘4’ and number ‘7’ is 4747.
Input : N = 99999999, A = 6, B = 7
Output : 6666677777
Below is the step by step algorithm to solve this problem:
- If the length of ‘N’ is odd then the resulting number will be of length ‘N+1’ as both ‘a’ and ‘b’ has to be in equal quantity.
- If the length of ‘N’ is even then the resulting number will either be of length ‘N’ or ‘N+2’.
- We will generate the number recursively by appending both A and B one by one and take the minimum of the two for the next recursive call.
- At last return the smallest number greater than or equal to ‘N’.
Below is the implementation of the above idea:
C++
// C++ program to find next greater Number// than N with the same quantity of// digits A and B#include <bits/stdc++.h>using namespace std;// Recursive function to find the required numberlong findNumUtil(long res, int a, int aCount, int b, int bCount, int n){ if (res > 1e11) return 1e11; // If the resulting number is >= n and // count of a = count of b, return the number if (aCount == bCount && res >= n) return res; // select minimum of two and call the function again return min(findNumUtil(res * 10 + a, a, aCount + 1, b, bCount, n), findNumUtil(res * 10 + b, a, aCount, b, bCount + 1, n));}// Function to find the number next greater Number// than N with the same quantity of// digits A and Bint findNum(int n, int a, int b){ int result = 0; int aCount = 0; int bCount = 0; return findNumUtil(result, a, aCount, b, bCount, n);}// Driver codeint main(){ int N = 4500; int A = 4; int B = 7; cout << findNum(N, A, B); return 0;} |
Java
// Java program to find next greater Number// than N with the same quantity of// digits A and Bpublic class GFG { // Recursive function to find the required number static long findNumUtil(long res, int a, int aCount, int b, int bCount, int n) { if (res > 1e11) return (long) 1e11; // If the resulting number is >= n and // count of a = count of b, return the number if (aCount == bCount && res >= n) return res; // select minimum of two and call the function again return Math.min(findNumUtil(res * 10 + a, a, aCount + 1, b, bCount, n), findNumUtil(res * 10 + b, a, aCount, b, bCount + 1, n)); } // Function to find the number next greater Number // than N with the same quantity of // digits A and B static int findNum(int n, int a, int b) { int result = 0; int aCount = 0; int bCount = 0; return (int) findNumUtil(result, a, aCount, b, bCount, n); } // Driver code public static void main(String args[]) { int N = 4500; int A = 4; int B = 7; System.out.println(findNum(N, A, B)); } // This Code is contributed by ANKITRAI1} |
Python3
# Python 3 program to find next greater # Number than N with the same quantity of# digits A and B# Recursive function to find the# required numberdef findNumUtil(res, a, aCount, b, bCount, n): if (res > 1e11): return 1e11 # If the resulting number is >= n # and count of a = count of b, # return the number if (aCount == bCount and res >= n): return res # select minimum of two and call # the function again return min(findNumUtil(res * 10 + a, a, aCount + 1, b, bCount, n), findNumUtil(res * 10 + b, a, aCount, b, bCount + 1, n))# Function to find the number next # greater Number than N with the # same quantity of digits A and Bdef findNum(n, a, b): result = 0 aCount = 0 bCount = 0 return findNumUtil(result, a, aCount, b, bCount, n)# Driver codeif __name__ == '__main__': N = 4500 A = 4 B = 7 print(findNum(N, A, B))# This code is contributed by# Sanjit_Prasad |
C#
// C# program to find next greater Number// than N with the same quantity of// digits A and Busing System;class GFG{// Recursive function to find the required numberstatic long findNumUtil(long res, int a, int aCount, int b, int bCount, int n){ if (res > 1e11) return (long) 1e11; // If the resulting number is >= n and // count of a = count of b, return the number if (aCount == bCount && res >= n) return res; // select minimum of two and call // the function again return Math.Min(findNumUtil(res * 10 + a, a, aCount + 1, b, bCount, n), findNumUtil(res * 10 + b, a, aCount, b, bCount + 1, n));}// Function to find the number next // greater Number than N with the // same quantity of digits A and Bstatic int findNum(int n, int a, int b){ int result = 0; int aCount = 0; int bCount = 0; return (int) findNumUtil(result, a, aCount, b, bCount, n);}// Driver codepublic static void Main(){ int N = 4500; int A = 4; int B = 7; Console.WriteLine(findNum(N, A, B));}}// This code is contributed by Shashank |
PHP
<?php// PHP program to find next greater Number// than N with the same quantity of// digits A and B// Recursive function to find the required numberfunction findNumUtil($res, $a, $aCount, $b, $bCount, $n){ if ($res > 100000000000) return 10000000000; // If the resulting number is >= n and // count of a = count of b, return the number if ($aCount == $bCount && $res >= $n) return $res; // select minimum of two and call the function again return min(findNumUtil($res * 10 + $a, $a, $aCount + 1, $b, $bCount, $n), findNumUtil($res * 10 + $b, $a, $aCount, $b, $bCount + 1, $n));}// Function to find the number next greater Number// than N with the same quantity of// digits A and Bfunction findNum($n, $a, $b){ $result = 0; $aCount = 0; $bCount = 0; return findNumUtil($result, $a, $aCount, $b, $bCount, $n);}// Driver code $N = 4500; $A = 4; $B = 7; echo findNum($N, $A, $B);// This Code is contributed by mits?> |
Javascript
<script> // Javascript program to find next greater Number // than N with the same quantity of // digits A and B // Recursive function to find the required number function findNumUtil(res, a, aCount, b, bCount, n) { if (res > 1e11) return 1e11; // If the resulting number is >= n and // count of a = count of b, return the number if (aCount == bCount && res >= n) return res; // select minimum of two and call // the function again return Math.min(findNumUtil(res * 10 + a, a, aCount + 1, b, bCount, n), findNumUtil(res * 10 + b, a, aCount, b, bCount + 1, n)); } // Function to find the number next // greater Number than N with the // same quantity of digits A and B function findNum(n, a, b) { let result = 0; let aCount = 0; let bCount = 0; return findNumUtil(result, a, aCount, b, bCount, n); } let N = 4500; let A = 4; let B = 7; document.write(findNum(N, A, B)); // This code is contributed by divyesh072019.</script> |
Output:
4747
Time Complexity: O(2n)
Auxiliary Space: O(1)
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