Find the minimum distance between the given two words

Given a list of words followed by two words, the task is to find the minimum distance between the given two words in the list of words.

Examples:

Input: S = { “the”, “quick”, “brown”, “fox”, “quick”}, word1 = “the”, word2 = “fox”
Output: 3
Explanation: Minimum distance between the words “the” and “fox” is 3

Input: S = {“geeks”, “for”, “geeks”, “contribute”,  “practice”}, word1 = “geeks”, word2 = “practice”
Output: 2
Explanation: Minimum distance between the words “geeks” and “practice” is 2

Approach: Follow the steps to solve this problem:

• Initialize the variables d1 = -1, d2 = -1 and ans = INT_MAX.
• Traverse the string and check:
  • If, s[i] is word1 then update d1 = i.
  • If, s[i] is word2 then update d2 = i.
  • If, d1 != -1 and d2 != -1, then update ans = min(ans, abs(d1-d2)).
• After traversing the string, return ans.

Below is the implementation of the above approach.

C++

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function to return shortest distance
int shortestDistance(vector<string>& s,
                     string word1, string word2)
{
    int d1 = -1, d2 = -1;
    int ans = INT_MAX;

    // Traverse the string
    for (int i = 0; i < s.size(); i++) {
        if (s[i] == word1)
            d1 = i;
        if (s[i] == word2)
            d2 = i;
        if (d1 != -1 && d2 != -1)
            ans = min(ans, abs(d1 - d2));
    }

    // Return the answer
    return ans;
}

// Driver Code
int main()
{
    vector<string> S
        = { "the", "quick", "brown", "fox", "quick" };

    string word1 = "the", word2 = "fox";

    // Function Call
    cout << shortestDistance(S, word1, word2);

    return 0;
}

Java

// Java code to implement the approach

import java.io.*;

class GFG {
    
// Function to return shortest distance
static int shortestDistance(String[] s,
                    String word1, String word2)
{
    int d1 = -1, d2 = -1;
    int ans = Integer.MAX_VALUE;

    // Traverse the string
    for (int i = 0; i < s.length; i++) {
        if (s[i] == word1)
            d1 = i;
        if (s[i] == word2)
            d2 = i;
        if (d1 != -1 && d2 != -1)
            ans = Math.min(ans, Math.abs(d1 - d2));
    }

    // Return the answer
    return ans;
}

 // Driver Code
    public static void main (String[] args) {
        String[] S
        = { "the", "quick", "brown", "fox", "quick" };

    String word1 = "the", word2 = "fox";

    // Function Call
    System.out.println(shortestDistance(S, word1, word2));
    }
}

// This code is contributed by Pushpesh Raj.

Python3

# Python3 code to implement the approach

# Function to return shortest distance
def shortestDistance(s, word1, word2) :

    d1 = -1; d2 = -1;
    ans = 100000000;

    # Traverse the string
    for i in range(len(s)) :
        if (s[i] == word1) :
            d1 = i;
        if (s[i] == word2) :
            d2 = i;
        if (d1 != -1 and d2 != -1) :
            ans = min(ans, abs(d1 - d2));

    # Return the answer
    return ans;

# Driver Code
if __name__ == "__main__" :

    S = [ "the", "quick", "brown", "fox", "quick" ];

    word1 = "the"; word2 = "fox";

    # Function Call
    print(shortestDistance(S, word1, word2));

   # This code is contributed by AnkThon

C#

// C# code to implement the approach
using System;

public class GFG {
    // Function to return shortest distance
    static int shortestDistance(string[] s, string word1,
                                string word2)
    {
        int d1 = -1, d2 = -1;
        int ans = Int32.MaxValue;

        // Traverse the string
        for (int i = 0; i < s.Length; i++) {
            if (s[i] == word1)
                d1 = i;
            if (s[i] == word2)
                d2 = i;
            if (d1 != -1 && d2 != -1)
                ans = Math.Min(ans, Math.Abs(d1 - d2));
        }

        // Return the answer
        return ans;
    }

    // Driver Code
    static public void Main()
    {
        String[] S
            = { "the", "quick", "brown", "fox", "quick" };

        String word1 = "the", word2 = "fox";

        // Function Call
        Console.WriteLine(
            shortestDistance(S, word1, word2));
    }
}

// This code is contributed by Rohit Pradhan

Javascript

// Javascript program for above approach

// Function to return shortest distance
function shortestDistance(s, word1, word2)
{
    let d1 = -1, d2 = -1;
    let ans = Number.MAX_VALUE;

    // Traverse the string
    for (let i = 0; i < s.length; i++) {
        if (s[i] == word1)
            d1 = i;
        if (s[i] == word2)
            d2 = i;
        if (d1 != -1 && d2 != -1)
            ans = Math.min(ans, Math.abs(d1 - d2));
    }

    // Return the answer
    return ans;
}

// Driver Code
        let S
        = [ "the", "quick", "brown", "fox", "quick" ];

    let word1 = "the", word2 = "fox";

    // Function Call
    console.log(shortestDistance(S, word1, word2));

// This code is contributed by sanjoy_62.

PHP

<?php

// Function to return shortest distance
function shortestDistance($s, $word1, $word2)
{
    $d1 = -1;
    $d2 = -1;
    $ans = PHP_INT_MAX;

    // Traverse the string
    for ($i = 0; $i < count($s); $i++) {
        if ($s[$i] == $word1)
            $d1 = $i;
        if ($s[$i] == $word2)
            $d2 = $i;
        if ($d1 != -1 && $d2 != -1)
            $ans = min($ans, abs($d1 - $d2));
    }

    // Return the answer
    return $ans;
}

// Driver Code
$S = array ("the", "quick", "brown", "fox", "quick");

$word1 = "the";
$word2 = "fox";

// Function Call
echo shortestDistance($S, $word1, $word2);

// This code is contributed by Kanishka Gupta
?>

Output

3

Time Complexity: O(N)
Auxiliary Space: O(1)

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