# Minimum Distance Between Words of a String

Given a string s and two words w1 and w2 that are present in S. The task is to find the minimum distance between w1 and w2. Here distance is the number of steps or words between the first and the second word.

Examples:

Input : s = “geeks for geeks contribute practice”, w1 = “geeks”, w2 = “practice”
Output : 1
There is only one word between closest occurrences of w1 and w2.

Input : s = “the quick the brown quick brown the frog”, w1 = “quick”, w2 = “frog”
Output : 2

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

A simple approach is to consider every occurrence of w1. For every occurrence of w1, find the closest w2 and keep track of minimum distance.

## C++

 `// C++ program to find Minimum Distance  ` `// Between Words of a String ` `#include ` `#include ` `using` `namespace` `std; ` ` `  `// Function to implement split function ` `void` `split(``const` `string &s, ``char` `delimeter,  ` `                     ``vector &words) ` `{ ` `    ``string token; ` `    ``stringstream tokenStream(s); ` ` `  `    ``while` `(getline(tokenStream, token, delimeter)) ` `        ``words.push_back(token); ` `} ` ` `  `// Function to calculate the minimum  ` `// distance between w1 and w2 in s ` `int` `distance(string s, string w1, string w2) ` `{ ` `    ``if` `(w1 == w2) ` `        ``return` `0; ` ` `  `    ``// get individual words in a list ` `    ``vector words; ` `    ``split(s, ``' '``, words); ` ` `  `    ``// assume total length of the string  ` `    ``// as minimum distance ` `    ``int` `min_dist = words.size() + 1; ` ` `  `    ``// traverse through the entire string ` `    ``for` `(``int` `index = 0; index < words.size(); index++) ` `    ``{ ` `        ``if` `(words[index] == w1) ` `        ``{ ` `            ``for` `(``int` `search = 0;  ` `                     ``search < words.size(); search++) ` `            ``{ ` `                ``if` `(words[search] == w2) ` `                ``{ ` `                    ``// the distance between the words is  ` `                    ``// the index of the first word - the  ` `                    ``// current word index ` `                    ``int` `curr = ``abs``(index - search) - 1; ` ` `  `                    ``// comparing current distance with  ` `                    ``// the previously assumed distance ` `                    ``if` `(curr < min_dist) ` `                        ``min_dist = curr; ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// w1 and w2 are same and adjacent ` `    ``return` `min_dist; ` `} ` ` `  `// Driver Code ` `int` `main(``int` `argc, ``char` `const` `*argv[]) ` `{ ` `    ``string s = ``"geeks for geeks contribute practice"``; ` `    ``string w1 = ``"geeks"``; ` `    ``string w2 = ``"practice"``; ` `    ``cout << distance(s, w1, w2) << endl; ` `    ``return` `0; ` `} ` ` `  `// This code is contributed by ` `// sanjeev2552 `

## Java

 `// Java program to find Minimum Distance  ` `// Between Words of a String ` ` ``class` `solution ` ` ``{ ` ` `  `// Function to calculate the minimum  ` `// distance between w1 and w2 in s  ` `static` `int` `distance(String s,String w1,String w2) ` `{ ` `      `  `    ``if` `(w1 .equals( w2) ) ` `        ``return` `0` `; ` `  `  `    ``// get individual words in a list  ` `    ``String words[] = s.split(``" "``); ` `  `  `    ``// assume total length of the string  ` `    ``// as minimum distance  ` `    ``int` `min_dist = (words.length) + ``1``; ` `  `  `    ``// traverse through the entire string  ` `    ``for` `(``int` `index = ``0``; ` `         ``index < words.length ; index ++) ` `    ``{ ` `  `  `        ``if` `(words[index] .equals( w1)) ` `        ``{ ` `            ``for` `(``int` `search = ``0``;  ` `                 ``search < words.length; search ++)  ` `            ``{ ` `                ``if` `(words[search] .equals(w2))  ` `                ``{ ` `                    ``// the distance between the words is  ` `                    ``// the index of the first word - the  ` `                    ``// current word index  ` `                    ``int` `curr = Math.abs(index - search) - ``1``;  ` `  `  `                    ``// comparing current distance with  ` `                    ``// the previously assumed distance  ` `                    ``if` `(curr < min_dist)  ` `                    ``{ ` `                        ``min_dist = curr ; ` `                    ``} ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` `      `  `    ``// w1 and w2 are same and adjacent  ` `    ``return` `min_dist; ` `} ` `  `  `// Driver code  ` `public` `static` `void` `main(String args[]) ` `{ ` `     `  `String s = ``"geeks for geeks contribute practice"``; ` `String w1 = ``"geeks"` `; ` `String w2 = ``"practice"` `; ` `  `  `System.out.print( distance(s, w1, w2) ); ` `  `  `} ` `} ` `//contributed by Arnab Kundu `

## Python3

 `# function to calculate the minimum  ` `# distance between w1 and w2 in s ` `  `  `def` `distance(s, w1, w2):  ` `     `  `    ``if` `w1 ``=``=` `w2 : ` `       ``return` `0` ` `  `    ``# get individual words in a list ` `    ``words ``=` `s.split(``" "``) ` ` `  `    ``# assume total length of the string as ` `    ``# minimum distance ` `    ``min_dist ``=` `len``(words)``+``1`  ` `  `    ``# traverse through the entire string ` `    ``for` `index ``in` `range``(``len``(words)): ` ` `  `        ``if` `words[index] ``=``=` `w1: ` `            ``for` `search ``in` `range``(``len``(words)): ` ` `  `                ``if` `words[search] ``=``=` `w2:  ` ` `  `                    ``# the distance between the words is ` `                    ``# the index of the first word - the  ` `                    ``# current word index  ` `                    ``curr ``=` `abs``(index ``-` `search) ``-` `1``; ` ` `  `                    ``# comparing current distance with  ` `                    ``# the previously assumed distance ` `                    ``if` `curr < min_dist: ` `                       ``min_dist ``=` `curr ` ` `  `    ``# w1 and w2 are same and adjacent ` `    ``return` `min_dist ` `     `  ` `  `# Driver code ` `s ``=` `"geeks for geeks contribute practice"` `w1 ``=` `"geeks"` `w2 ``=` `"practice"` `print``(distance(s, w1, w2)) `

## C#

 `// C# program to find Minimum Distance  ` `// Between Words of a String ` ` `  `using` `System; ` ` ``class` `solution ` ` ``{ ` `  `  `// Function to calculate the minimum  ` `// distance between w1 and w2 in s  ` `static` `int` `distance(``string` `s,``string` `w1,``string` `w2) ` `{ ` `       `  `    ``if` `(w1 .Equals( w2) ) ` `        ``return` `0 ; ` `   `  `    ``// get individual words in a list  ` `    ``string``[] words = s.Split(``" "``); ` `   `  `    ``// assume total length of the string  ` `    ``// as minimum distance  ` `    ``int` `min_dist = (words.Length) + 1; ` `   `  `    ``// traverse through the entire string  ` `    ``for` `(``int` `index = 0; ` `         ``index < words.Length ; index ++) ` `    ``{ ` `   `  `        ``if` `(words[index] .Equals( w1)) ` `        ``{ ` `            ``for` `(``int` `search = 0;  ` `                 ``search < words.Length; search ++)  ` `            ``{ ` `                ``if` `(words[search] .Equals(w2))  ` `                ``{ ` `                    ``// the distance between the words is  ` `                    ``// the index of the first word - the  ` `                    ``// current word index  ` `                    ``int` `curr = Math.Abs(index - search) - 1;  ` `   `  `                    ``// comparing current distance with  ` `                    ``// the previously assumed distance  ` `                    ``if` `(curr < min_dist)  ` `                    ``{ ` `                        ``min_dist = curr ; ` `                    ``} ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` `       `  `    ``// w1 and w2 are same and adjacent  ` `    ``return` `min_dist; ` `} ` `   `  `// Driver code  ` `public` `static` `void` `Main() ` `{ ` `      `  `string` `s = ``"geeks for geeks contribute practice"``; ` `string` `w1 = ``"geeks"` `; ` `string` `w2 = ``"practice"` `; ` `   `  `Console.Write( distance(s, w1, w2) ); ` `   `  `} ` `} `

## PHP

 ` `

Output:

```1
```

An efficient solution is to find first occurrence of any element, then keep track of previous element and current element, if they are different and distance is less than current minimum, update minimum.

## Java

 `// Java program to extract words from ` `// a strung using stringstream ` `class` `GFG { ` ` `  `    ``static` `int` `distance(String s, String w1, String w2) { ` ` `  `        ``if` `(w1.equals(w2)) { ` `            ``return` `0``; ` `        ``} ` ` `  `       ``// get individual words in a list ` `        ``String[] words = s.split(``" "``); ` `        ``int` `n = words.length; ` ` `  `        ``// assume total length of the string as ` `        ``// minimum distance ` `        ``int` `min_dist = n + ``1``; ` ` `  `        ``// Find the first occurrence of any of the two  ` `        ``// numbers (w1 or w2) and store the index of ` `        ``// this occurrence in prev  ` `        ``int` `prev = ``0``, i = ``0``; ` `        ``for` `(i = ``0``; i < n; i++) { ` ` `  `            ``if` `(words[i].equals(w1) || words[i].equals(w2)) { ` `                ``prev = i; ` `                ``break``; ` `            ``} ` `        ``} ` `        ``// Traverse after the first occurrence  ` `        ``while` `(i < n) { ` `            ``if` `(words[i].equals(w1) || words[i].equals(w2)) { ` ` `  `                ``// If the current element matches with ` `                ``// any of the two then check if current  ` `                ``// element and prev element are different  ` `                ``// Also check if this value is smaller than ` `                ``// minimum distance so far  ` `                ``if` `((!words[prev].equals(words[i])) && (i - prev) < min_dist) { ` `                    ``min_dist = i - prev - ``1``; ` `                    ``prev = i; ` `                ``} ``else` `{ ` `                    ``prev = i; ` `                ``} ` ` `  `            ``} ` `            ``i += ``1``; ` ` `  `        ``} ` `        ``return` `min_dist; ` `    ``} ` `// Driver code ` ` `  `    ``public` `static` `void` `main(String[] args) { ` `        ``String s = ``"geeks for geeks contribute practice"``; ` `        ``String w1 = ``"geeks"``; ` `        ``String w2 = ``"practice"``; ` `        ``System.out.println(distance(s, w1, w2)); ` `// This code is contributed by princiRaj1992 ` `    ``} ` `} `

## C#

 `// C# program to extract words from  ` `// a strung using stringstream  ` `using` `System; ` ` `  `class` `GFG  ` `{  ` ` `  `    ``static` `int` `distance(String s, String w1, String w2)  ` `    ``{  ` ` `  `        ``if` `(w1.Equals(w2)) ` `        ``{  ` `            ``return` `0;  ` `        ``}  ` ` `  `        ``// get individual words in a list  ` `        ``String[] words = s.Split(``" "``);  ` `        ``int` `n = words.Length;  ` ` `  `        ``// assume total length of the string as  ` `        ``// minimum distance  ` `        ``int` `min_dist = n + 1;  ` ` `  `        ``// Find the first occurrence of any of the two  ` `        ``// numbers (w1 or w2) and store the index of  ` `        ``// this occurrence in prev  ` `        ``int` `prev = 0, i = 0;  ` `        ``for` `(i = 0; i < n; i++) ` `        ``{  ` ` `  `            ``if` `(words[i].Equals(w1) || words[i].Equals(w2))  ` `            ``{  ` `                ``prev = i;  ` `                ``break``;  ` `            ``}  ` `        ``}  ` `         `  `        ``// Traverse after the first occurrence  ` `        ``while` `(i < n)  ` `        ``{  ` `            ``if` `(words[i].Equals(w1) || words[i].Equals(w2))  ` `            ``{  ` ` `  `                ``// If the current element matches with  ` `                ``// any of the two then check if current  ` `                ``// element and prev element are different  ` `                ``// Also check if this value is smaller than  ` `                ``// minimum distance so far  ` `                ``if` `((!words[prev].Equals(words[i])) &&  ` `                                ``(i - prev) < min_dist)  ` `                ``{  ` `                    ``min_dist = i - prev - 1;  ` `                    ``prev = i;  ` `                ``}  ` `                ``else`  `                ``{  ` `                    ``prev = i;  ` `                ``}  ` `            ``}  ` `            ``i += 1;  ` ` `  `        ``}  ` `        ``return` `min_dist;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main(String[] args)  ` `    ``{  ` `        ``String s = ``"geeks for geeks contribute practice"``;  ` `        ``String w1 = ``"geeks"``;  ` `        ``String w2 = ``"practice"``;  ` `        ``Console.Write(distance(s, w1, w2));  ` `    ``}  ` `}  ` ` `  `// This code is contributed by Mohit kumar 29 `

## Python3

 `# Python3 program to extract words from ` `# a strung using stringstream ` ` `  `def` `distance(s, w1, w2):  ` `     `  `    ``if` `w1 ``=``=` `w2 : ` `       ``return` `0` ` `  `    ``# get individual words in a list ` `    ``words ``=` `s.split(``" "``) ` `    ``n ``=` `len``(words) ` ` `  `    ``# assume total length of the string as ` `    ``# minimum distance ` `    ``min_dist ``=` `n``+``1`  ` `  `    ``# Find the first occurrence of any of the two  ` `    ``# numbers (w1 or w2) and store the index of ` `    ``# this occurrence in prev  ` `    ``for` `i ``in` `range``(n):  ` `         `  `        ``if` `words[i] ``=``=` `w1 ``or` `words[i] ``=``=` `w2:  ` `            ``prev ``=` `i  ` `            ``break` ` `  `    ``# Traverse after the first occurrence  ` `    ``while` `i < n:  ` `        ``if` `words[i] ``=``=` `w1 ``or` `words[i] ``=``=` `w2:  ` ` `  `            ``# If the current element matches with ` `            ``# any of the two then check if current  ` `            ``# element and prev element are different  ` `            ``# Also check if this value is smaller than ` `            ``# minimum distance so far  ` `            ``if` `words[prev] !``=` `words[i] ``and` `(i ``-` `prev) < min_dist :  ` `                ``min_dist ``=` `i ``-` `prev ``-` `1` `                ``prev ``=` `i  ` `            ``else``:  ` `                ``prev ``=` `i  ` `        ``i ``+``=` `1`         ` `  `    ``return` `min_dist  ` ` `  `# Driver code ` `s ``=` `"geeks for geeks contribute practice"` `w1 ``=` `"geeks"` `w2 ``=` `"practice"` `print``(distance(s, w1, w2)) `

Output:

```1
```

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.