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 the closest occurrences of w1 and w2.Input : s = “the quick the brown quick brown the frog”, w1 = “quick”, w2 = “frog”
Output : 2
A simple approach is to consider every occurrence of w1. For every occurrence of w1, find the closest w2 and keep track of the minimum distance.
Implementation:
// C++ program to find Minimum Distance // Between Words of a String #include <bits/stdc++.h> #include <sstream> using namespace std;
// Function to implement split function void split( const string &s, char delimiter,
vector<string> &words)
{ string token;
stringstream tokenStream(s);
while (getline(tokenStream, token, delimiter))
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<string> 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 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 |
# 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# 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 // PHP program to find Minimum Distance // Between Words of a String // Function to calculate the minimum // distance between w1 and w2 in s function distance( $s , $w1 , $w2 )
{ if ( $w1 == $w2 )
return 0 ;
// get individual words in a list
$words = explode ( " " , $s );
// assume total length of the string
// as minimum distance
$min_dist = sizeof( $words ) + 1;
// traverse through the entire string
for ( $index = 0;
$index < sizeof( $words ) ; $index ++)
{
if ( $words [ $index ] == $w1 )
{
for ( $search = 0;
$search < sizeof( $words ); $search ++)
{
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" ;
echo distance( $s , $w1 , $w2 ) ;
// This code is contributed by Ryuga ?> |
<script> // Javascript program to find Minimum Distance // Between Words of a String // Function to calculate the minimum // distance between w1 and w2 in s function distance(s,w1,w2) {
if (w1 ==( w2) )
return 0 ;
// get individual words in a list
let words = s.split( " " );
// assume total length of the string
// as minimum distance
let min_dist = (words.length) + 1;
// traverse through the entire string
for (let index = 0;
index < words.length ; index ++)
{
if (words[index] == ( w1))
{
for (let search = 0;
search < words.length; search ++)
{
if (words[search] == (w2))
{
// the distance between the words is
// the index of the first word - the
// current word index
let 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
let s = "geeks for geeks contribute practice" ;
let w1 = "geeks" ;
let w2 = "practice" ;
document.write( distance(s, w1, w2) );
// This code is contributed by rag2127 </script> |
1
Complexity Analysis:
- Time Complexity: O(n2)
- Auxiliary Space: O(n)
An efficient solution is to find the first occurrence of any element, then keep track of the previous element and current element. If they are different and the distance is less than the current minimum, update the minimum.
Implementation:
// C++ program to extract words from // a string using stringstream #include<bits/stdc++.h> using namespace std;
int distance(string s, string w1, string w2)
{ if (w1 == w2)
{
return 0;
}
vector<string> words;
// Used to split string around spaces.
istringstream ss(s);
string word; // for storing each word
// Traverse through all words
// while loop till we get
// strings to store in string word
while (ss >> word)
{
words.push_back(word);
}
int n = words.size();
// 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] == w1 || (words[i] == w2))
{
prev = i;
break ;
}
}
// Traverse after the first occurrence
while (i < n)
{
if (words[i] == w1 || (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]) &&
(i - prev) < min_dist)
{
min_dist = i - prev - 1;
prev = i;
}
else
{
prev = i;
}
}
i += 1;
}
return min_dist;
} // Driver code int main()
{ string s = "geeks for geeks contribute practice" ;
string w1 = "geeks" ;
string w2 = "practice" ;
cout<<distance(s, w1, w2);
} // This code is contributed by rutvik_56. |
// Java program to extract words from // a string 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# program to extract words from // a string 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 program to extract words from # a string 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))
|
<script> // Javascript program to extract words from // a string using stringstream function distance(s,w1,w2)
{ if (w1 == (w2))
{
return 0;
}
// get individual words in a list
let words = s.split( " " );
let n = words.length;
// assume total length of the string as
// minimum distance
let 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
let prev = 0, i = 0;
for (i = 0; i < n; i++)
{
if (words[i] == (w1) || words[i] == (w2))
{
prev = i;
break ;
}
}
// Traverse after the first occurrence
while (i < n)
{
if (words[i] == (w1) || 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])) && (i - prev) < min_dist) {
min_dist = i - prev - 1;
prev = i;
} else {
prev = i;
}
}
i += 1;
}
return min_dist;
} // Driver code let s = "geeks for geeks contribute practice" ;
let w1 = "geeks" ;
let w2 = "practice" ;
document.write(distance(s, w1, w2)); // This code is contributed by avanitrachhadiya2155 </script> |
1
Complexity Analysis:
- Time Complexity: O(n)
- Auxiliary Space: O(1)
An efficient solution is to store the index of word1 in (lastpos) variable if word1 occur again then we update (lastpos) if word1 not occur then simply find the difference of index of word1 and word2.
Implementation:
// C++ program to find Minimum Distance // Between Words of a String #include <bits/stdc++.h> using namespace std;
int shortestDistance(vector<string> &s, string word1, string word2)
{
if (word1==word2) return 0;
int ans = INT_MAX;
//To store the lastposition of word1
int lastPos = -1;
for ( int i = 0 ; i < s.size() ; i++)
{
if (s[i] == word1 || s[i] == word2)
{
//first occurrence of word1
if (lastPos == -1)
lastPos = i;
else
{
//if word1 repeated again we store the last position of word1
if (s[lastPos]==s[i])
lastPos = i;
else
{
//find the difference of position of word1 and word2
ans = min(ans , (i-lastPos)-1);
lastPos = i;
}
}
}
}
return ans;
} //Driver code int main() {
vector<string> s{ "geeks" , "for" , "geeks" , "contribute" ,
"practice" };
string w1 = "geeks" ;
string w2 = "practice" ;
cout<<shortestDistance(s, w1, w2)<< "\n" ;
return 0;
} |
import java.util.ArrayList;
class Demo {
static int shortestDistance(ArrayList<String> list,
String word1, String word2)
{
if (word1 == word2)
return 0 ;
int ans = Integer.MAX_VALUE;
// To store the lastposition of word1
int lastPos = - 1 ;
for ( int i = 0 ; i < list.size(); i++) {
if (list.get(i) == word1
|| list.get(i) == word2) {
// first occurrence of word1
if (lastPos == - 1 )
lastPos = i;
else {
// if word1 repeated again we store the
// last position of word1
if (list.get(lastPos) == list.get(i))
lastPos = i;
else {
// find the difference of position
// of word1 and word2
ans = Math.min(ans,
((i - lastPos) - 1 ));
lastPos = i;
}
}
}
}
return ans;
}
public static void main(String arg[])
{
ArrayList<String> list = new ArrayList<>();
list.add( "geeks" );
list.add( "for" );
list.add( "geeks" );
list.add( "contribute" );
list.add( "practice" );
String w1 = "geeks" ;
String w2 = "practice" ;
System.out.println(shortestDistance(list, w1, w2));
}
} // This code is contributed by nmkiniqw7b. |
# Python program to find Minimum Distance # Between Words of a String def shortestDistance(s, word1, word2):
if (word1 = = word2):
return 0
ans = 1e9 + 7
# To store the lastposition of word1
lastPos = - 1
for i in range ( 0 , len (s)):
if (s[i] = = word1 or s[i] = = word2):
# first occurrence of word1
if (lastPos = = - 1 ):
lastPos = i
else :
# if word1 repeated again we store the last position of word1
if (s[lastPos] = = s[i]):
lastPos = i
else :
# find the difference of position of word1 and word2
ans = min (ans, (i - lastPos) - 1 )
lastPos = i
return ans
# Driver code s = [ "geeks" , "for" , "geeks" , "contribute" , "practice" ]
w1 = "geeks"
w2 = "practice"
print (shortestDistance(s, w1, w2))
# This code is contributed by Samim Hossain Mondal. |
// C# program to find Minimum Distance // Between Words of a String using System;
using System.Collections.Generic;
public class GFG {
static int shortestDistance(List< string > list,
string word1, string word2)
{
if (word1 == word2)
return 0;
int ans = Int32.MaxValue;
// To store the lastposition of word1
int lastPos = -1;
for ( int i = 0; i < list.Count; i++) {
if (list[i] == word1 || list[i] == word2) {
// first occurrence of word1
if (lastPos == -1)
lastPos = i;
else {
// if word1 repeated again we store the
// last position of word1
if (list[lastPos] == list[i])
lastPos = i;
else {
// find the difference of position
// of word1 and word2
ans = Math.Min(ans,
((i - lastPos) - 1));
lastPos = i;
}
}
}
}
return ans;
}
public static void Main( string [] arg)
{
List< string > list = new List< string >();
list.Add( "geeks" );
list.Add( "for" );
list.Add( "geeks" );
list.Add( "contribute" );
list.Add( "practice" );
string w1 = "geeks" ;
string w2 = "practice" ;
Console.WriteLine(shortestDistance(list, w1, w2));
}
} // This code is contributed by karandeep1234. |
// javascript program to find Minimum Distance // Between Words of a String function shortestDistance(s, word1, word2)
{
if (word1==word2)
return 0;
let ans = Number.MAX_SAFE_INTEGER;
//To store the lastposition of word1
let lastPos = -1;
for (let i = 0 ; i < s.length ; i++)
{
if (s[i] == word1 || s[i] == word2)
{
//first occurrence of word1
if (lastPos == -1)
lastPos = i;
else
{
//if word1 repeated again we store the last position of word1
if (s[lastPos]==s[i])
lastPos = i;
else
{
//find the difference of position of word1 and word2
ans = Math.min(ans , (i-lastPos)-1);
lastPos = i;
}
}
}
}
return ans;
} // Driver code let s=[ "geeks" , "for" , "geeks" , "contribute" , "practice" ];
let w1 = "geeks" ;
let w2 = "practice" ;
console.log(shortestDistance(s, w1, w2));
// This code is contributed by garg28harsh.
|
1
Complexity Analysis:
- Time Complexity: O(n)
- Auxiliary Space: O(1)