Given a string 
( containing characters from ‘0’ to ‘9’) and two digits 
and 
. The task is to find the substring in the given string with maximum occurrences and containing a’s and b’s only. If there are two or more such substrings with same frequencies then print the lexicographically smallest. If there does not exists any such substring then print -1.
Examples:
Input : str = "47", a = 4, b = 7 Output : 4 Input : str = "23", a = 4, b = 7 Output : -1
The idea is to observe that we need to find the substring with maximum number of occurrences. So, if we consider substrings that contains both a’s and b’s then the number of occurrences will be less than if we consider the substrings with single digits ‘a’ and ‘b’ individually.
So, the idea is to calculate the frequency of digits of ‘a’ and ‘b’ in the string and the one with maximum frequency will be the answer.
Note: If both digits have same frequency then the digit which is lexicographically smaller among ‘a’ and ‘b’ will be the answer.
Below is the implementation of the above approach:
C++
// CPP program to Find the lexicographically// smallest substring in a given string with// maximum frequency and contains a's and b's only#include <bits/stdc++.h>using namespace std;// Function to Find the lexicographically// smallest substring in a given string with// maximum frequency and contains a's and b's only.int maxFreq(string s, int a, int b){ // To store frequency of digits int fre[10] = { 0 }; // size of string int n = s.size(); // Take lexicographically larger digit in b if (a > b) swap(a, b); // get frequency of each character for (int i = 0; i < n; i++) fre[s[i] - '0']++; // If no such string exits if (fre[a] == 0 and fre[b] == 0) return -1; // Maximum frequency else if (fre[a] >= fre[b]) return a; else return b;}// Driver programint main(){ int a = 4, b = 7; string s = "47744"; cout << maxFreq(s, a, b); return 0;} |
Java
// Java program to Find the lexicographically// smallest substring in a given string with// maximum frequency and contains a's and b's onlyimport java.io.*;class GFG {// Function to Find the lexicographically// smallest substring in a given string with// maximum frequency and contains a's and b's only.static int maxFreq(String s, int a, int b){ // To store frequency of digits int fre[] = new int[10]; // size of string int n = s.length(); // Take lexicographically larger digit in b if (a > b) { int temp = a; a =b; b = temp; } // get frequency of each character for (int i = 0; i < n; i++) fre[s.charAt(i) - '0']++; // If no such string exits if (fre[a] == 0 && fre[b] == 0) return -1; // Maximum frequency else if (fre[a] >= fre[b]) return a; else return b;}// Driver program public static void main (String[] args) { int a = 4, b = 7; String s = "47744"; System.out.print(maxFreq(s, a, b)); }}// This code is contributed by inder_verma |
Python3
# Python 3 program to Find the lexicographically# smallest substring in a given string with# maximum frequency and contains a's and b's only# Function to Find the lexicographically# smallest substring in a given string with# maximum frequency and contains a's and b's only.def maxFreq(s, a, b): # To store frequency of digits fre = [0 for i in range(10)] # size of string n = len(s) # Take lexicographically larger digit in b if (a > b): swap(a, b) # get frequency of each character for i in range(0,n,1): a = ord(s[i]) - ord('0') fre[a] += 1 # If no such string exits if (fre[a] == 0 and fre[b] == 0): return -1 # Maximum frequency elif (fre[a] >= fre[b]): return a else: return b# Driver programif __name__ == '__main__': a = 4 b = 7 s = "47744" print(maxFreq(s, a, b))# This code is contributed by# Surendra_Gangwar |
C#
// C# program to Find the lexicographically// smallest substring in a given string with// maximum frequency and contains a's and b's onlyusing System;class GFG {// Function to Find the lexicographically// smallest substring in a given string with// maximum frequency and contains a's and b's only.static int maxFreq(string s, int a, int b){ // To store frequency of digits int []fre = new int[10]; // size of string int n = s.Length; // Take lexicographically larger digit in b if (a > b) { int temp = a; a =b; b = temp; } // get frequency of each character for (int i = 0; i < n; i++) fre[s[i] - '0']++; // If no such string exits if (fre[a] == 0 && fre[b] == 0) return -1; // Maximum frequency else if (fre[a] >= fre[b]) return a; else return b;}// Driver program public static void Main () { int a = 4, b = 7; string s = "47744"; Console.WriteLine(maxFreq(s, a, b)); }}// This code is contributed by inder_verma |
PHP
<?php// PHP program to Find the lexicographically// smallest substring in a given string with// maximum frequency and contains a's and b's only// Function to Find the lexicographically// smallest substring in a given string with// maximum frequency and contains a's and b's only.function maxFreq($s, $a, $b){ // To store frequency of digits $fre = array_fill(0, 10, 0); // size of string $n = strlen($s); // Take lexicographically larger digit in b if ($a > $b) { $xx = $a; $a = $b; $b = $xx;} // get frequency of each character for ($i = 0; $i < $n; $i++) { $a = ord($s[$i]) - ord('0'); $fre[$a] += 1; } // If no such string exits if ($fre[$a] == 0 and $fre[$b] == 0) return -1; // Maximum frequency else if ($fre[$a] >= $fre[$b]) return $a; else return $b;}// Driver Code$a = 4;$b = 7;$s = "47744";print(maxFreq($s, $a, $b));// This code is contributed by mits?> |
Javascript
<script> // JavaScript program to Find the lexicographically // smallest substring in a given string with // maximum frequency and contains a's and b's only // Function to Find the lexicographically // smallest substring in a given string with // maximum frequency and contains a's and b's only. function maxFreq(s, a, b) { // To store frequency of digits var fre = new Array(10).fill(0); // size of string var n = s.length; // Take lexicographically larger digit in b if (a > b) { var temp = a; a = b; b = temp; } // get frequency of each character for (var i = 0; i < n; i++) fre[s[i].charCodeAt(0) - "0".charCodeAt(0)]++; // If no such string exits if (fre[a] === 0 && fre[b] === 0) return -1; // Maximum frequency else if (fre[a] >= fre[b]) return a; else return b; } // Driver program var a = 4, b = 7; var s = "47744"; document.write(maxFreq(s, a, b)); // This code is contributed by rdtank. </script> |
Output
4
Complexity Analysis:
- Time Complexity: O(n), where n is the length of the string s.
- Auxiliary Space: O(10)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!