Minimum distance between duplicates in a String

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: 0 
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 i^th 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 <bits/stdc++.h>
using namespace std;

// This function is used to find
// minimum distance between same
// repeating characters
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] 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 Code
int 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 approach
import java.io.*;

class GFG{

// This function is used to find
// minimum distance between same
// repeating characters
static 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 code
public 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 Input
S = "geeksforgeeks"
N = 13

# Function Call
print(shortestDistance(S, N))

C#

// C# program for the above approach
using System;

class GFG{

// This function is used to find
// minimum distance between same
// repeating characters
static 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 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 shivanisinghss2110

Javascript

<script>
// JavaScript program for the above approach
// This function is used to find
// minimum distance between same
// repeating characters
function 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 = "geeksforgeeks";
    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 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 characters
int 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 = "geeksforgeeks";
  int n = 13;

  // Function call
  cout << (shortestDistance(s, n));
}

// This code is contributed by Aditya Kumar (adityakumar129)

Java

// Java implementation of above approach
import java.io.*;
import java.util.*;

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
    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 code
public 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 characters
def 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 Input
S = "geeksforgeeks"
N = 13

# Function Call
print(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<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 = "geeksforgeeks";
        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 = "geeksforgeeks";
        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 technique
int 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 = "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
// technique
import 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 = "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
# technique
import sys

# This function is used to find
# minimum distance between any two
# repeating characters
# using two - pointers and hashing technique
def 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 code
if __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
// technique
using 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 = "geeksforgeeks";
        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 = "geeksforgeeks";
        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)