Given a string S of size N consisting of lowercase alphabets only, the task is to find the smallest character having a minimum sum of distances between its consecutive repetition. If string S consists of distinct characters only, then print “-1”.
Examples:
Input: str = “aabbaadc”
Output: b;
Explanation:
For all the characters in the given string, the sum of the required distances are as follows:
Indices of ‘a’ = {0, 1, 4, 5}
=> Sum of the distances of its next repetition = abs(0 – 1) + abs(4 – 1) + abs(5 – 4) = 5
Indices of ‘b’ = {2, 3}
=> Sum of the distances of its next repetition = abs(2 – 3) = 1
‘c’, ‘d’ has no repetition
From the above distances the minimum sum of distance obtained is 1, for the character ‘b’.
Therefore, the required answer is ‘b’.Input: str = “abcdef”
Output: -1
Explanation:
All the characters in the given string are distinct.
Naive approach: The simplest approach is to traverse the given string and for each character, find the sum of the shortest distances individually. Print the smallest character with the minimum shortest distance.
Time Complexity: O(N*26), where N is the length of the given string.
Auxiliary Space: O(N)
Efficient Approach: The idea is to traverse the string once and find the first and last indices for every character as the sum of the difference between the index between the same characters is the difference between the first and the last character. Follow the below steps to solve the problem:
- Create the arrays last[] and first[] having the length equals 26 initialize both the arrays as -1.
- Initialize min with some large number.
- Traverse the string S and update the first occurrence of the current characters to the current index if it equals to -1.
- Find the last occurrence of each character and store it in the array last[].
- Traverse the array and update the index having a minimum difference at each corresponding if both the index has non-negative value.
- If the minimum index is found at index x, and print the character (x + ‘a’).
- Otherwise, print “-1”.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the character// repeats with minimum distancechar minDistChar(string s){ int n = s.length(); // Stores the first and last index int* first = new int[26]; int* last = new int[26]; // Initialize with -1 for (int i = 0; i < 26; i++) { first[i] = -1; last[i] = -1; } // Get the values of last and // first occurrence for (int i = 0; i < n; i++) { // Update the first index if (first[s[i] - 'a'] == -1) { first[s[i] - 'a'] = i; } // Update the last index last[s[i] - 'a'] = i; } // Initialize min int min = INT_MAX; char ans = '1'; // Get the minimum for (int i = 0; i < 26; i++) { // Values must not be same if (last[i] == first[i]) continue; // Update the minimum distance if (min > last[i] - first[i]) { min = last[i] - first[i]; ans = i + 'a'; } } // return ans return ans;}// Driver Codeint main(){ string str = "neveropen"; // Function Call cout << minDistChar(str); return 0;} |
Java
// Java program for // the above approachimport java.util.*;class GFG{// Function to find the character// repeats with minimum distancestatic char minDistChar(char []s){int n = s.length;// Stores the first and last indexint []first = new int[26];int []last = new int[26];// Initialize with -1for (int i = 0; i < 26; i++) { first[i] = -1; last[i] = -1;}// Get the values of last and// first occurrencefor (int i = 0; i < n; i++) { // Update the first index if (first[s[i] - 'a'] == -1) { first[s[i] - 'a'] = i; } // Update the last index last[s[i] - 'a'] = i;}// Initialize minint min = Integer.MAX_VALUE;char ans = '1';// Get the minimumfor (int i = 0; i < 26; i++) { // Values must not be same if (last[i] == first[i]) continue; // Update the minimum distance if (min > last[i] - first[i]) { min = last[i] - first[i]; ans = (char) (i + 'a'); }}// return ansreturn ans;}// Driver Codepublic static void main(String[] args){String str = "neveropen";// Function CallSystem.out.print(minDistChar(str.toCharArray()));}}// This code is contributed by shikhasingrajput |
Python3
# Python3 program for the above approachimport sys# Function to find the character# repeats with minimum distancedef minDistChar(s): n = len(s) # Stores the first and last index first = [] last = [] # Initialize with -1 for i in range(26): first.append(-1) last.append(-1) # Get the values of last and # first occurrence for i in range(n): # Update the first index if (first[ord(s[i]) - ord('a')] == -1): first[ord(s[i]) - ord('a')] = i # Update the last index last[ord(s[i]) - ord('a')] = i # Initialize the min min = sys.maxsize ans = '1' # Get the minimum for i in range(26): # Values must not be same if (last[i] == first[i]): continue # Update the minimum distance if (min > last[i] - first[i]): min = last[i] - first[i] ans = i + ord('a') return chr(ans) # Driver Codeif __name__ == "__main__": str = "neveropen" # Function call print(minDistChar(str))# This code is contributed by dadi madhav |
C#
// C# program for the above approach using System; using System.Collections.Generic; class GFG{ // Function to find the character// repeats with minimum distancestatic char minDistChar(char []s){ int n = s.Length; // Stores the first and last index int []first = new int[26]; int []last = new int[26]; // Initialize with -1 for(int i = 0; i < 26; i++) { first[i] = -1; last[i] = -1; } // Get the values of last and // first occurrence for(int i = 0; i < n; i++) { // Update the first index if (first[s[i] - 'a'] == -1) { first[s[i] - 'a'] = i; } // Update the last index last[s[i] - 'a'] = i; } // Initialize min int min = int.MaxValue; char ans = '1'; // Get the minimum for(int i = 0; i < 26; i++) { // Values must not be same if (last[i] == first[i]) continue; // Update the minimum distance if (min > last[i] - first[i]) { min = last[i] - first[i]; ans = (char)(i + 'a'); } } // return ans return ans;} // Driver Code public static void Main(string[] args) { String str = "neveropen"; // Function call Console.Write(minDistChar(str.ToCharArray()));} } // This code is contributed by rutvik_56 |
Javascript
<script>// JavaScript program for the above approach // Function to find the let acter// repeats with minimum distancefunction minDistChar(s){ let n = s.length; // Stores the first and last index let first = new Array(26); let last = new Array(26); // Initialize with -1 for(let i = 0; i < 26; i++) { first[i] = -1; last[i] = -1; } // Get the values of last and // first occurrence for(let i = 0; i < n; i++) { // Update the first index if (first[s[i] - 'a'] == -1) { first[s[i] - 'a'] = i; } // Update the last index last[s[i] - 'a'] = i; } // Initialize min let min = 100000; var ans = 'g'; // Get the minimum for(let i = 0; i < 26; i++) { // Values must not be same if (last[i] == first[i]) continue; // Update the minimum distance if (min > last[i] - first[i]) { min = last[i] - first[i]; ans = String.fromCharCode(i + 97); } } // return ans return ans;} // Driver Code str = "neveropen";// Function calldocument.write(minDistChar(str));</script> |
g
Time Complexity: O(N)
Auxiliary Space: O(N)
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