Given a string S and a character X where
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++ 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 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 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# 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 |
<script> // JavaScript implementation of above approach // Function to return required // array of distances function shortestDistance(S, X)
{ // Find distance from occurrences of X
// appearing before current character.
let prev = Number.MAX_VALUE;
let ans = [];
for (let i = 0; i < S.length; i++)
{
if (S[i] == X)
prev = i;
if (prev == Number.MAX_VALUE)
ans.push(Number.MAX_VALUE);
else
ans.push(i - prev);
}
// Find distance from occurrences of X
// appearing after current character and
// compare this distance with earlier.
prev = Number.MAX_VALUE;
for (let i = S.length - 1; i >= 0; i--)
{
if (S[i] == X)
prev = i;
if (prev != Number.MAX_VALUE)
ans[i] = Math.min(ans[i], prev - i);
}
for (let val of ans)
document.write(val + ' ' );
} // Driver code let S = "helloworld" ;
let X = 'o' ;
shortestDistance(S, X); // This code is contributed by Shinjanpatra </script> |
Output
4 3 2 1 0 1 0 1 2 3
Complexity Analysis:
- Time Complexity: O(|S|),
- Auxiliary Space: O(|S|) because extra space for vector ans is being used
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 between 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++ 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 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 |
// Java implementation of above approach import java.util.*;
class GFG
{ // Function to return required
// vector of distances
static ArrayList<Integer> shortestToChar(String s,
char c)
{
// list to hold position of c in s
ArrayList<Integer> list = new ArrayList<Integer>();
// list to hold the result
ArrayList<Integer> res = new ArrayList<Integer>();
// length of string
int len = s.length();
// Iterate over string to create list
for ( int i = 0 ; i < len; i++) {
if (s.charAt(i) == c) {
list.add(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.get(p1);
v2 = list.get(p2);
// Current Index is before than p1
if (i <= v1) {
res.add(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.add(i - v1);
}
// Current Index is nearer to p2
else {
res.add(v2 - i);
// Move pointer 1 step ahead
p1 = p2;
p2 = p2 < l ? (p2 + 1 ) : p2;
}
}
// Current index is after p2
else {
res.add(i - v2);
}
}
return res;
}
// Driver code
public static void main(String[] args)
{
String s = "geeksforgeeks" ;
char c = 'e' ;
ArrayList<Integer> res = shortestToChar(s, c);
for (Integer i : res)
System.out.print(i + " " );
}
} // This code is contributed by phasing17 |
// C# implementation of above approach using System;
using System.Collections.Generic;
class GFG {
// Function to return required
// vector of distances
static List< int > shortestToChar( string s, char c)
{
// list to hold position of c in s
List< int > list = new List< int >();
// list to hold the result
List< int > res = new List< int >();
// 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.Add(i);
}
}
int p1, p2, v1, v2;
// max value of p2
int l = list.Count - 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.Add(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.Add(i - v1);
}
// Current Index is nearer to p2
else {
res.Add(v2 - i);
// Move pointer 1 step ahead
p1 = p2;
p2 = p2 < l ? (p2 + 1) : p2;
}
}
// Current index is after p2
else {
res.Add(i - v2);
}
}
return res;
}
// Driver code
public static void Main( string [] args)
{
string s = "geeksforgeeks" ;
char c = 'e' ;
List< int > res = shortestToChar(s, c);
foreach ( var i in res) Console.Write(i + " " );
}
} // This code is contributed by phasing17 |
# Python implementation of above approach # Function to return required # vector of distances def shortestToChar(s,c):
# list to hold position of c in s
list = []
# list to hold the result
res = []
# length of string
Len = len (s)
# Iterate over string to create list
for i in range ( Len ):
if (s[i] = = c):
list .append(i)
# max value of p2
l = len ( list ) - 1
# Initialize the pointers
p1 = 0
p2 = 1 if l > 0 else 0
# Create result array
for i in range ( Len ):
# Values at current pointers
v1 = list [p1]
v2 = list [p2]
# Current Index is before than p1
if (i < = v1):
res.append(v1 - i)
# Current Index is between p1 and p2
elif (i < = v2):
# Current Index is nearer to p1
if (i - v1 < v2 - i):
res.append(i - v1)
# Current Index is nearer to p2
else :
res.append(v2 - i)
# Move pointer 1 step ahead
p1 = p2
p2 = (p2 + 1 ) if (p2<l) else p2
# Current index is after p2
else :
res.append(i - v2)
return res
# Driver code s = "geeksforgeeks"
c = 'e'
res = shortestToChar(s, c)
for i in res:
print (i,end = " " )
# This code is contributed by shinjanpatra |
<script> // JavaScript implementation of above approach // Function to return required // vector of distances function shortestToChar(s,c)
{ // list to hold position of c in s
let list = [];
// list to hold the result
let res = [];
// length of string
let len = s.length;
// Iterate over string to create list
for (let i = 0; i < len; i++) {
if (s[i] == c) {
list.push(i);
}
}
let p1, p2, v1, v2;
// max value of p2
let l = list.length - 1;
// Initialize the pointers
p1 = 0;
p2 = l > 0 ? 1 : 0;
// Create result array
for (let 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(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(i - v1);
}
// Current Index is nearer to p2
else {
res.push(v2 - i);
// Move pointer 1 step ahead
p1 = p2;
p2 = p2 < l ? (p2 + 1) : p2;
}
}
// Current index is after p2
else {
res.push(i - v2);
}
}
return res;
} // Driver code let s = "geeksforgeeks" ;
let c = 'e' ;
let res = shortestToChar(s, c); for (let i of res)
document.write(i , end = " " );
// This code is contributed by Shinjanpatra </script> |
Output
1 0 0 1 2 3 3 2 1 0 0 1 2
Complexity Analysis:
- Time Complexity: O(n),
- Auxiliary Space: O(n)