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Shortest distance to every other character from given character

  • Difficulty Level : Medium
  • Last Updated : 08 Feb, 2021

Given a string S and a character X where X\varepsilon S[i]      , for some 0\leq i \leq S.length()-1      . 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[0] = ‘g’ nearest ‘e’ is at distance = 1 i.e. S[1] = ‘e’. 
similarly, for S[1] = ‘e’, distance = 0. 
for S[6] = ‘o’, distance = 3 since we have S[9] = ‘e’, and so on.
Input: S = “helloworld”, X = ‘o’ 
Output: [4, 3, 2, 1, 0, 1, 0, 1, 2, 3] 

Approach 1: 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 <bits/stdc++.h>
 
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<Integer> 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<S.Length; i++)
        {
            if (S[i] == X)
                prev = i;
            if (prev == int.MaxValue)
                ans.Add(int.MaxValue);
            else
                ans.Add(i-prev);
        }
         
        // Find distance from occurrences of X
        // appearing after current character and
        // compare this distance with earlier.
        prev = int.MaxValue;
        for (int i=S.Length-1; 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



4 3 2 1 0 1 0 1 2 3 

Approach 2: Create a list holding the occurrence of the character and then create two pointers pointing two immediate locations in this list, now iterate over the string to find the difference from these two pointers and insert the minimum in the result list. If pointer 2 is nearer to the current character, move the pointers one step ahead. 

  • Create a list holding positions of the required character in the string and an empty list to hold the result array.
  • Create two pointers to the list p1=0 and p2=0 if list length is 1 else p2=1
  • Iterate over the string and compare the values at these pointers (v1=p1->value & v2=p2->value) with the current index(i).
    • If i <= v1, then push v1-i in the result list.
    • Else if i <= v2
      • if i is nearer to v1, then push i-v1 in the result list
      • Else push v2-i in the result list and move pointer one step forward if possible
    • Else push i-v1 into the result list
  • Return result list

Below is the implementation of the above approach:

C++




// C++ implementation of above approach
#include <bits/stdc++.h>
 
using namespace std;
 
// Function to return required
// vector of distances
vector<int> shortestToChar(string s, char c)
{
    // list to hold position of c in s
    vector<int> list;
 
    // list to hold the result
    vector<int> res;
 
    // length of string
    int len = s.length();
 
    // Iterate over string to create list
    for (int i = 0; i < len; i++) {
        if (s[i] == c) {
            list.push_back(i);
        }
    }
 
    int p1, p2, v1, v2;
 
    // max value of p2
    int l = list.size() - 1;
 
    // Initialize the pointers
    p1 = 0;
    p2 = l > 0 ? 1 : 0;
 
    // Create result array
    for (int i = 0; i < len; i++) {
        // Values at current pointers
        v1 = list[p1];
        v2 = list[p2];
 
        // Current Index is before than p1
        if (i <= v1) {
            res.push_back(v1 - i);
        }
        // Current Index is between p1 and p2
        else if (i <= v2) {
            // Current Index is nearer to p1
            if (i - v1 < v2 - i) {
                res.push_back(i - v1);
            }
            // Current Index is nearer to p2
            else {
                res.push_back(v2 - i);
                // Move pointer 1 step ahead
                p1 = p2;
                p2 = p2 < l ? (p2 + 1) : p2;
            }
        }
        // Current index is after p2
        else {
            res.push_back(i - v2);
        }
    }
    return res;
}
 
// Driver code
int main()
{
    string s = "geeksforgeeks";
    char c = 'e';
    vector<int> res = shortestToChar(s, c);
    for (auto i : res)
        cout << i << "  ";
    return 0;
}
 
// This code is contributed by Shivam Sharma

C




// C implementation of above approach
#include <stdio.h>
#define MAX_SIZE 100
 
// Function to return required
// vector of distances
void shortestToChar(char s[], char c, int* res)
{
    // list to hold position of c in s
    int list[MAX_SIZE];
 
    // length of string
    int len = 0;
 
    // To hold size of list
    int l = 0;
 
    // Iterate over string to create list
    while (s[len] != '\0') {
        if (s[len] == c) {
            list[l] = len;
            l++;
        }
        len++;
    }
 
    int p1, p2, v1, v2;
 
    // max value of p2
    l = l - 1;
 
    // Initialize the pointers
    p1 = 0;
    p2 = l > 0 ? 1 : 0;
 
    // Create result array
    for (int i = 0; i < len; i++) {
        // Values at current pointers
        v1 = list[p1];
        v2 = list[p2];
 
        // Current Index is before than p1
        if (i <= v1) {
            res[i] = (v1 - i);
        }
        // Current Index is between p1 and p2
        else if (i <= v2) {
            // Current Index is nearer to p1
            if (i - v1 < v2 - i) {
                res[i] = (i - v1);
            }
            // Current Index is nearer to p2
            else {
                res[i] = (v2 - i);
                // Move pointer 1 step ahead
                p1 = p2;
                p2 = p2 < l ? (p2 + 1) : p2;
            }
        }
        // Current index is after p2
        else {
            res[i] = (i - v2);
        }
    }
}
 
// Driver code
int main()
{
    char s[] = "geeksforgeeks";
    char c = 'e';
    int res[MAX_SIZE];
    shortestToChar(s, c, res);
    int i = 0;
    while (s[i] != '\0')
        printf("%d  ", res[i++]);
    return 0;
}
 
// This code is contributed by Shivam Sharma
Output
1  0  0  1  2  3  3  2  1  0  0  1  2  

Time Complexity: O(n)

Space Complexity: O(n)

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