Minimum distance between duplicates in a String

• Difficulty Level : Medium
• Last Updated : 10 Jan, 2022

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 = “geeksforgeeks”, N = 13
Output:
Explanation
The repeating characters in string S = “geeksforgeeks” 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
• 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 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 = "geeksforgeeks";    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 = "geeksforgeeks";    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 characters  def 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 = "geeksforgeeks"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 = "geeksforgeeks";    int N = 13;       // Function call    Console.Write(shortestDistance(S, N));}} // This code is contributed by shivanisinghss2110

Javascript


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 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 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 m;  vector 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 - 1;} int main(){  string s;  s = "geeksforgeeks";  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 dic = new HashMap();    ArrayList 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 = "geeksforgeeks";    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 = "geeksforgeeks"N = 13 # Function Callprint(shortestDistance(S, N))

C#

 // C# implementation of above approach using 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 dic = new Dictionary();        List dis = new List();         // 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- 1;    }     // Driver code    public static void Main(String[] args) {         // Given input        String S = "geeksforgeeks";        int N = 13;         // Function call        Console.WriteLine(shortestDistance(S, N));    }} // This code contributed by umadevi9616

Javascript


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 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 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 = "geeksforgeeks";    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;        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 = "geeksforgeeks";        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 = "geeksforgeeks";    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;        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 = "geeksforgeeks";        int n = 13;         // Function Call        Console.Write(shortestDistance(s, n));    }} // This code is contributed by umadevi9616

Javascript


Output
0

Time Complexity: O(N)
Auxiliary Space: O(N)

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