# Shortest distance to every other character from given character

Given a string S and a character X where , for some . The task is to return an array of distances representing the shortest distance from the character X to every other character in the string.

Examples:

Input: S = “geeksforgeeks”, X = ‘e’
Output: [1, 0, 0, 1, 2, 3, 3, 2, 1, 0, 0, 1, 2]
for S = ‘g’ nearest ‘e’ is at distance = 1 i.e. S = ‘e’.
similarly, for S = ‘e’, distance = 0.
for S = ‘o’, distance = 3 since we have S = ‘e’, and so on.

Input: S = “helloworld”, X = ‘o’
Output: [4, 3, 2, 1, 0, 1, 0, 1, 2, 3]

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: For each character at index i in S[], let us try to find the distance to the next character X going left to right, and from right to left. The answer will be the minimum of these two values.

• When going from left to right, we remember the index of the last character X we’ve seen. Then the answer is i – prev.
• When going from right to left, the answer is prev – i.
• We take the minimum of these two answers to create our final distance array.
• Finally, print the array.

Below is the implementation of above approach:

## C++

 // C++ implementation of above approach   #include     using namespace std;     // Function to return required   // array of distances   void shortestDistance(string S, char X)  {       // Find distance from occurrences of X       // appearing before current character.       int prev = INT_MAX;      vector<int> ans;             for (int i = 0; i < S.length(); i++)      {           if (S[i] == X)               prev = i;           if (prev == INT_MAX)              ans.push_back(INT_MAX);          else             ans.push_back(i - prev);      }         // Find distance from occurrences of X       // appearing after current character and       // compare this distance with earlier.       prev = INT_MAX;       for (int i = S.length() - 1; i >= 0; i--)      {           if (S[i] == X)               prev = i;           if (prev != INT_MAX)              ans[i] = min(ans[i], prev - i);       }         for (auto val: ans)          cout << val << ' ';  }     // Driver code   int main()  {      string S = "helloworld";      char X = 'o';      shortestDistance(S, X);       return 0;  }     // This code is contributed by Rituraj Jain

## Java

 // Java implementation of above approach   import java.util.*;     class GFG  {     // Function to return required   // array of distances   static void shortestDistance(String S, char X)   {          // Find distance from occurrences of X       // appearing before current character.       int prev = Integer.MAX_VALUE;       Vector ans = new Vector<>();              for (int i = 0; i < S.length(); i++)       {           if (S.charAt(i) == X)               prev = i;           if (prev == Integer.MAX_VALUE)              ans.add(Integer.MAX_VALUE);          else                 ans.add(i - prev);       }          // Find distance from occurrences of X       // appearing after current character and       // compare this distance with earlier.       prev = Integer.MAX_VALUE;      for (int i = S.length() - 1; i >= 0; i--)       {           if (S.charAt(i) == X)               prev = i;           if (prev != Integer.MAX_VALUE)                  ans.set(i, Math.min(ans.get(i), prev - i));       }          for (Integer val: ans)               System.out.print(val+" ");  }      // Driver code   public static void main(String[] args)  {      String S = "geeksforgeeks";       char X = 'g';       shortestDistance(S, X);  }  }     // This code is contributed by Rajput-Ji

## Python3

 # Python3 implementation of above approach     # Function to return required   # array of distances  def shortestDistance(S, X):         # Find distance from occurrences of X      # appearing before current character.      inf = float('inf')      prev = inf      ans = []      for i,j in enumerate(S):          if S[i] == X:              prev = i          if (prev == inf) :               ans.append(inf)          else :                   ans.append(i - prev)            # Find distance from occurrences of X      # appearing after current character and      # compare this distance with earlier.      prev = inf      for i in range(len(S) - 1, -1, -1):          if S[i] == X:              prev = i          if (X != inf):                  ans[i] = min(ans[i], prev - i)         # return array of distance      return ans        # Driver code  S = "geeksforgeeks" X = "g"    # Function call to print answer  print(shortestDistance(S, X))

## C#

 // C# implementation of above approach  using System;  using System.Collections.Generic;     class GFG  {             // Function to return required       // array of distances      public static void shortestDistance(String S, char X){                     // Find distance from occurrences of X           // appearing before current character.           int prev = int.MaxValue;          List<int> ans = new List<int>();          for (int i=0; i=0; i--)          {              if (S[i] == X)                  prev = i;              if (prev != int.MaxValue)                  ans[i] = Math.Min(ans[i], prev-i);          }                     foreach (var i in ans)              Console.Write(i + " ");      }             // Driver code      public static void Main(String[] args)      {          String S = "geeksforgeeks";          char X = 'g';          shortestDistance(S, X);      }  }     // This code is contributed by  // sanjeev2552

Output:

[0, 1, 2, 3, 4, 3, 2, 1, 0, 1, 2, 3, 4]


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