Given a string S and its length N (provided N > 0). The task is to find the minimum distance between same repeating characters, if no repeating characters present in string S return -1.
Examples:
Input: S = “neveropen”, N = 13
Output: 0
Explanation:
The repeating characters in string S = “neveropen” with minimum distance is ‘e’.
The minimum difference of their indices is 0 (i.e. the character ‘e’ are present at index 1 and 2).Input: S = “abdfhbih”, N = 8
Output: 2
Explanation:
The repeating characters in string S = “abdfhbih” with minimum distance is ‘h’.
The minimum difference of their indices is 2 (i.e. the character ‘h’ are present at index 4 and 7).
Naive Approach: This problem can be solved using two nested loops, one considering an element at each index ‘i’ in string S, next loop will find the matching character same to ith in S.
First, store each difference between repeating characters in a variable and check whether this current distance is less than the previous value stored in same variable. At the end return the variable storing Minimum value. There is one corner case i.e. when there are no repeating characters return -1. Follow the steps below to solve this problem:
- Initialize a variable minDis as N to store the minimum distances of repeating characters.
- Iterate in the range [0, N-1] using the variable i:
- Iterate in the range [i + 1, N-1] using the variable j:
- If S[i] is equal to S[j] and distance between them is less than the minDis, update minDis and then break the loop
- Iterate in the range [i + 1, N-1] using the variable j:
- If minDis value is not updated that means no repeating characters, return -1, otherwise return minDis – 1.
Below is the implementation of above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// This function is used to find// minimum distance between same// repeating charactersint shortestDistance(string S, int N){ // Store minimum distance between same // repeating characters int minDis = S.length(); // For loop to consider each element // of string for(int i = 0; i < N; i++) { for(int j = i + 1; j < N; j++) { // Comparison of string characters and // updating the minDis value if (S[i] == S[j] and (j - i) < minDis) { minDis = j - i; // As this value would be least // therefore break break; } } } // If minDis value is not updated that means // no repeating characters if (minDis == S.length()) return -1; else // Minimum distance is minDis - 1 return minDis - 1;}// Driver Codeint main(){ // Given Input string S = "neveropen"; int N = 13; // Function Call cout << (shortestDistance(S, N));}// This code is contributed by lokeshpotta20 |
Java
// Java program for the above approachimport java.io.*;class GFG{ // This function is used to find// minimum distance between same// repeating charactersstatic int shortestDistance(String S, int N){ // Store minimum distance between same // repeating characters int minDis = S.length(); // For loop to consider each element // of string for(int i = 0; i < N; i++) { for(int j = i + 1; j < N; j++) { // Comparison of string characters and // updating the minDis value if (S.charAt(i) == S.charAt(j) && (j - i) < minDis) { minDis = j - i; // As this value would be least // therefore break break; } } } // If minDis value is not updated that means // no repeating characters if (minDis == S.length()) return -1; // Minimum distance is minDis - 1 else return minDis - 1;}// Driver codepublic static void main(String[] args){ // Given input String S = "neveropen"; int N = 13; // Function call System.out.println(shortestDistance(S, N));}}// This code is contributed by MuskanKalra1 |
Python3
# Python3 implementation of above approach# This function is used to find# minimum distance between same# repeating charactersdef shortestDistance(S, N): # Store minimum distance between same # repeating characters minDis = len(S) # For loop to consider each element of string for i in range(N): for j in range(i+1, N): # Comparison of string characters and # updating the minDis value if(S[i] == S[j] and (j-i) < minDis): minDis = j-i # As this value would be least therefore break break # If minDis value is not updated that means # no repeating characters if(minDis == len(S)): return -1 else: # Minimum distance is minDis - 1 return minDis - 1# Driver Code# Given InputS = "neveropen"N = 13# Function Callprint(shortestDistance(S, N)) |
C#
// C# program for the above approachusing System;class GFG{ // This function is used to find// minimum distance between same// repeating charactersstatic int shortestDistance(string S, int N){ // Store minimum distance between same // repeating characters int minDis = S.Length; // For loop to consider each element // of string for(int i = 0; i < N; i++) { for(int j = i + 1; j < N; j++) { // Comparison of string characters and // updating the minDis value if (S[i] == S[j] && (j - i) < minDis) { minDis = j - i; // As this value would be least // therefore break break; } } } // If minDis value is not updated that means // no repeating characters if (minDis == S.Length) return -1; // Minimum distance is minDis - 1 else return minDis - 1;}// Driver codepublic static void Main(String[] args){ // Given input string S = "neveropen"; int N = 13; // Function call Console.Write(shortestDistance(S, N));}}// This code is contributed by shivanisinghss2110 |
Javascript
<script>// JavaScript program for the above approach// This function is used to find// minimum distance between same// repeating charactersfunction shortestDistance( S, N){ // Store minimum distance between same // repeating characters var minDis = S.length; // For loop to consider each element // of string for(var i = 0; i < N; i++) { for(var j = i + 1; j < N; j++) { // Comparison of string characters and // updating the minDis value if (S.charAt(i) == S.charAt(j) && (j - i) < minDis) { minDis = j - i; // As this value would be least // therefore break break; } } } // If minDis value is not updated that means // no repeating characters if (minDis == S.length) return -1; // Minimum distance is minDis - 1 else return minDis - 1;}// Driver code// Given input var S = "neveropen"; var N = 13; // Function call document.write(shortestDistance(S, N));// This code is contributed by shivanisinghss2110</script> |
Output
0
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: This problem can be solved by using Dictionary or Hashing. First, store the last index against the character of dictionary so that it can be subtracted with the last value stored against the same character in dictionary and further store the distance in the list. At the end return the minimum of the list. Follow the steps below to solve this problem:
- Initialize a dictionary dic for holding the last occurrence of character and a list dis to store distance.
- Iterate in the range [0, N-1] using the variable i:
- If character present in dictionary:
- Then, extract its last value dic[S[i]] and update it with current position i.
- Store the difference in a variable var = i – dic[S[i]] and append it to list dis.
- If the character is not present, initialize with the current position.
- If character present in dictionary:
- If the length of dis is 0 that means no repeating characters, return -1, otherwise return min(dis) – 1.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>using namespace std;// This function is used to find minimum distance between// same repeating charactersint shortestDistance(string s, int n){ // Define a map and an vector map<char, int> m; vector<int> v; // Temporary variable int var; // Traverse through string for (int i = 0; i < n; i++) { // If character present in map if (m.find(s[i]) != m.end()) { // Difference between current position and last // stored value var = i - m[s[i]]; // Updating current position m[s[i]] = i; // Storing difference in list v.push_back(var); } // If character not in map assign it with // initial of its position else m[s[i]] = i; } // If no element inserted in vector // i.e. no repeating characters if (v.size() == 0) return -1; sort(v.begin(), v.end()); return v[0] - 1;}int main(){ string s; s = "neveropen"; int n = 13; // Function call cout << (shortestDistance(s, n));}// This code is contributed by Aditya Kumar (adityakumar129) |
Java
// Java implementation of above approachimport java.io.*;import java.util.*;class GFG{// This function is used to find// minimum distance between same// repeating charactersstatic int shortestDistance(String S, int N){ // Define a hashmap and an arraylist HashMap<Character, Integer> dic = new HashMap<Character, Integer>(); ArrayList<Integer> dis = new ArrayList<>(); // Temporary variable int var; // Traverse through string for(int i = 0; i < N; i++) { // If character present in dictionary if (dic.get(S.charAt(i)) != null) { // Difference between current position // and last stored value var = i - dic.get(S.charAt(i)); // Updating current position dic.put(S.charAt(i), i); // Storing difference in list dis.add(var); } // If character not in dictionary assign // it with initial of its position else { dic.put(S.charAt(i), i); } } // If no element inserted in list // i.e. no repeating characterss if (dis.size() == 0) return -1; // Return minimum distance else return Collections.min(dis) - 1;}// Driver codepublic static void main(String[] args){ // Given input String S = "neveropen"; int N = 13; // Function call System.out.println(shortestDistance(S, N));}}// This code is contributed by MuskanKalra1 |
Python3
# Python3 implementation of above approach# This function is used to find the# required the minimum distances of# repeating charactersdef shortestDistance(S, N): # Define dictionary and list dic = {} dis = [] # Traverse through string for i in range(N): # If character present in dictionary if S[i] in dic: # Difference between current position # and last stored value var = i- dic[S[i]] # Updating current position dic[S[i]] = i # Storing difference in list dis.append(var) # If character not in dictionary assign # it with initial of its position else: dic[S[i]] = i # If no element inserted in list # i.e. no repeating characters if(len(dis) == 0): return -1 # Return minimum distance else: return min(dis)-1# Driver code# Given InputS = "neveropen"N = 13# Function Callprint(shortestDistance(S, N)) |
C#
// C# implementation of above approachusing System;using System.Collections.Generic;public class GFG { // This function is used to find // minimum distance between same // repeating characters static int shortestDistance(String S, int N) { // Define a hashmap and an arraylist Dictionary<char, int> dic = new Dictionary<char, int>(); List<int> dis = new List<int>(); // Temporary variable int var; // Traverse through string for (int i = 0; i < N; i++) { // If character present in dictionary if (dic.ContainsKey(S[i])) { // Difference between current position // and last stored value var = i - dic[S[i]]; // Updating current position dic[S[i]]= i; // Storing difference in list dis.Add(var); } // If character not in dictionary assign // it with initial of its position else { dic.Add(S[i], i); } }dis.Sort(); // If no element inserted in list // i.e. no repeating characterss if (dis.Count == 0) return -1; // Return minimum distance else return dis[0]- 1; } // Driver code public static void Main(String[] args) { // Given input String S = "neveropen"; int N = 13; // Function call Console.WriteLine(shortestDistance(S, N)); }}// This code contributed by umadevi9616 |
Javascript
<script>// javascript implementation of above approach// This function is used to find // minimum distance between same // repeating characters function shortestDistance( S , N) { // Define a hashmap and an arraylist var dic = new Map(); var dis = new Array(); // Temporary variable var var1; // Traverse through string for (i = 0; i < N; i++) { // If character present in dictionary if (dic[S[i]] != null) { // Difference between current position // and last stored value var1 = i - dic[S[i]]; // Updating current position dic[S[i]] = i; // Storing difference in list dis.push(var1); } // If character not in dictionary assign // it with initial of its position else { dic[S[i]]= i; } } // If no element inserted in list // i.e. no repeating characterss if (dis.length == 0) return -1; // Return minimum distance else return dis.reduce(function(previous,current){ return previous < current ? previous:current }) - 1; } // Driver code // Given input var S = "neveropen"; var N = 13; // Function call document.write(shortestDistance(S, N));// This code is contributed by gauravrajput1 </script> |
Output
0
Time Complexity: O(N)
Auxiliary Space: O(N)
Alternate Solution: The following problem could also be solved using an improved two-pointers approach. The idea basically is to maintain a left-pointer for every character and as soon as that particular character is repeated, the left pointer points to the nearest index of the character. While doing this, we can maintain a variable ans that will store the minimum distance between any two duplicate characters. This could be achieved using a visited vector array that will store a current character’s nearest index in the array.
Follow the steps below to solve this problem:
- Initialize a visited vector for storing the last index of any character (left pointer)
- Iterate in the range [0, N-1] :
- If the character is previously visited:
- Find the distance between the characters and check, if the distance between the two is minimum.
- If it’s less than the previous minimum, update its value.
- Update the current character’s last index in the visited array.
If there is no minimum distance obtained(Ii.e., when the value of ans is INT_MAX) that means there are no repeating characters. In this case return -1;
Below is the implementation of the above approach:
C++
// C++ Program to find the minimum distance between two// repeating characters in a string using two pointers technique#include <bits/stdc++.h>using namespace std;// This function is used to find// minimum distance between any two// repeating characters// using two - pointers and hashing techniqueint shortestDistance(string s, int n) { // hash array to store character's last index vector<int> visited(128, -1); int ans = INT_MAX; // Traverse through the string for(int right = 0; right < n; right++) { char c = s[right]; int left = visited; // If the character is present in visited array // find if its forming minimum distance if(left != -1) ans = min(ans, right - left -1); // update current character's last index visited = right; } // Return minimum distance found, else -1 return ans == INT_MAX ? -1 : ans;}int main(){ // Given Input string s = "neveropen"; int n = 13; // Function Call cout << (shortestDistance(s, n));} |
Java
// Java Program to find the minimum distance between two// repeating characters in a String using two pointers// techniqueimport java.util.*;class GFG { // This function is used to find // minimum distance between any two // repeating characters // using two - pointers and hashing technique static int shortestDistance(String s, int n) { // hash array to store character's last index int[] visited = new int[128]; Arrays.fill(visited, -1); int ans = Integer.MAX_VALUE; // Traverse through the String for (int right = 0; right < n; right++) { char c = s.charAt(right); int left = visited; // If the character is present in visited array // find if its forming minimum distance if (left != -1) ans = Math.min(ans, right - left - 1); // update current character's last index visited = right; } // Return minimum distance found, else -1 return ans == Integer.MAX_VALUE ? -1 : ans; } // Driver code public static void main(String[] args) { // Given Input String s = "neveropen"; int n = 13; // Function Call System.out.print(shortestDistance(s, n)); }}// This code is contributed by umadevi9616 |
Python3
# Python Program to find the minimum distance between two# repeating characters in a String using two pointers# techniqueimport sys# This function is used to find# minimum distance between any two# repeating characters# using two - pointers and hashing techniquedef shortestDistance(s, n): # hash array to store character's last index visited = [-1 for i in range(128)]; ans = sys.maxsize; # Traverse through the String for right in range(n): c = (s[right]); left = visited[ord(c)]; # If the character is present in visited array # find if its forming minimum distance if (left != -1): ans = min(ans, right - left - 1); # update current character's last index visited[ord(c)] = right; # Return minimum distance found, else -1 if(ans == sys.maxsize): return -1; else: return ans;# Driver codeif __name__ == '__main__': # Given Input s = "neveropen"; n = 13; # Function Call print(shortestDistance(s, n));# This code is contributed by umadevi9616 |
C#
// C# Program to find the minimum distance between two// repeating characters in a String using two pointers// techniqueusing System;public class GFG { // This function is used to find // minimum distance between any two // repeating characters // using two - pointers and hashing technique static int shortestDistance(string s, int n) { // hash array to store character's last index int[] visited = new int[128]; for(int i = 0; i < 128; i++) visited[i] = -1; int ans = int.MaxValue; // Traverse through the String for (int right = 0; right < n; right++) { char c = s[right]; int left = visited; // If the character is present in visited array // find if its forming minimum distance if (left != -1) ans = Math.Min(ans, right - left - 1); // update current character's last index visited = right; } // Return minimum distance found, else -1 return ans == int.MaxValue ? -1 : ans; } // Driver code public static void Main(String[] args) { // Given Input string s = "neveropen"; int n = 13; // Function Call Console.Write(shortestDistance(s, n)); }}// This code is contributed by umadevi9616 |
Javascript
<script>// javascript Program to find the minimum distance between two// repeating characters in a String using two pointers// technique // This function is used to find // minimum distance between any two // repeating characters // using two - pointers and hashing technique function shortestDistance(s , n) { // hash array to store character's last index var visited = Array(128).fill(-1); var ans = Number.MAX_VALUE; // Traverse through the String for (var right = 0; right < n; right++) { var left = visited[s.charCodeAt(right)]; // If the character is present in visited array // find if its forming minimum distance if (left != -1) ans = Math.min(ans, right - left - 1); // update current character's last index visited[s.charCodeAt(right)] = right; } // Return minimum distance found, else -1 return ans == Number.MAX_VALUE ? -1 : ans; } // Driver code // Given Input var s = "neveropen"; var n = 13; // Function Call document.write(shortestDistance(s, n));// This code is contributed by umadevi9616 </script> |
Output
0
Time Complexity: O(N)
Auxiliary Space: O(N)
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