what is the total distance of all characters (letters) from the beginning of a word to the end?

Given a string Southward and a character X where, for some. The task is to return an assortment of distances representing the shortest distance from the character X to every other character in the string.
Examples:

Input: S = "geeksforgeeks", X = 'due east'
Output: [1, 0, 0, i, ii, 3, 3, 2, 1, 0, 0, 1, two]
for S[0] = 'g' nearest 'e' is at distance = ane i.eastward. Due south[one] = 'east'.
similarly, for S[1] = 'east', distance = 0.
for Southward[six] = 'o', distance = iii since we have S[9] = 'due east', and so on.
Input: S = "helloworld", Ten = 'o'
Output: [four, 3, 2, one, 0, 1, 0, ane, 2, 3]

Approach one: For each grapheme at index i in S[], permit u.s. try to find the distance to the next graphic symbol X going left to right, and from right to left. The answer volition be the minimum of these ii values.

• When going from left to right, nosotros remember the index of the last graphic symbol 10 nosotros've seen. And then the answer is i – prev.
• When going from right to left, the answer is prev – i.
• We accept the minimum of these two answers to create our final distance assortment.
• Finally, impress the array.

Below is the implementation of above approach:

C++

#include <bits/stdc++.h>

using namespace std;

void shortestDistance(string S, char X)

{

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);

}

prev = INT_MAX;

for ( int i = South.length() - 1; i >= 0; i--)

{

if (S[i] == Ten)

prev = i;

if (prev != INT_MAX)

ans[i] = min(ans[i], prev - i);

}

for ( auto val: ans)

cout << val << ' ' ;

}

int main()

{

cord S = "helloworld" ;

char X = 'o' ;

shortestDistance(S, X);

return 0;

}

Java

import java.util.*;

form GFG

{

static void shortestDistance(Cord S, char 10)

{

int prev = Integer.MAX_VALUE;

Vector<Integer> ans = new Vector<>();

for ( int i = 0 ; i < South.length(); i++)

{

if (S.charAt(i) == X)

prev = i;

if (prev == Integer.MAX_VALUE)

ans.add(Integer.MAX_VALUE);

else

ans.add(i - prev);

}

prev = Integer.MAX_VALUE;

for ( int i = S.length() - 1 ; i >= 0 ; i--)

{

if (S.charAt(i) == 10)

prev = i;

if (prev != Integer.MAX_VALUE)

ans.set(i, Math.min(ans.get(i), prev - i));

}

for (Integer val: ans)

Organization.out.impress(val+ " " );

}

public static void main(String[] args)

{

Cord S = "geeksforgeeks" ;

char X = 'g' ;

shortestDistance(S, X);

}

}

Python3

def shortestDistance(S, X):

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)

prev = inf

for i in range ( len (S) - 1 , - 1 , - 1 ):

if Due south[i] = = X:

prev = i

if (Ten ! = inf):

ans[i] = min (ans[i], prev - i)

return ans

S = "geeksforgeeks"

X = "g"

impress (shortestDistance(Southward, X))

C#

using System;

using Arrangement.Collections.Generic;

grade GFG

{

public static void shortestDistance(String S, char X){

int prev = int .MaxValue;

Listing< int > ans = new Listing< int >();

for ( int i=0; i<S.Length; i++)

{

if (South[i] == X)

prev = i;

if (prev == int .MaxValue)

ans.Add( int .MaxValue);

else

ans.Add(i-prev);

}

prev = int .MaxValue;

for ( int i=S.Length-i; i>=0; i--)

{

if (S[i] == 10)

prev = i;

if (prev != int .MaxValue)

ans[i] = Math.Min(ans[i], prev-i);

}

foreach ( var i in ans)

Panel.Write(i + " " );

}

public static void Main(Cord[] args)

{

String South = "geeksforgeeks" ;

char Ten = 'g' ;

shortestDistance(S, X);

}

}

Javascript

<script>

role shortestDistance(South, X)

{

let prev = Number.MAX_VALUE;

allow 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.button(i - prev);

}

prev = Number.MAX_VALUE;

for (permit i = S.length - 1; i >= 0; i--)

{

if (S[i] == Ten)

prev = i;

if (prev != Number.MAX_VALUE)

ans[i] = Math.min(ans[i], prev - i);

}

for (let val of ans)

document.write(val + ' ' );

}

let S = "helloworld" ;

let X = 'o' ;

shortestDistance(S, X);

</script>

Output

4 3 ii 1 0 1 0 i 2 3          

Approach 2: Create a list holding the occurrence of the graphic symbol and and then create ii pointers pointing two immediate locations in this list, now iterate over the string to observe the deviation from these 2 pointers and insert the minimum in the result listing. If pointer 2 is nearer to the current graphic symbol, motility the pointers one step alee.

• Create a list property positions of the required graphic symbol in the cord and an empty list to concord the effect array.
• Create two pointers to the list p1=0 and p2=0 if list length is i else p2=i
• Iterate over the string and compare the values at these pointers (v1=p1->value & v2=p2->value) with the current alphabetize(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 effect list
    • Else push v2-i in the result list and move pointer one pace forrard if possible
  • Else push button i-v1 into the upshot list
• Render result list

Below is the implementation of the to a higher place approach:

C++

#include <bits/stdc++.h>

using namespace std;

vector< int > shortestToChar(string s, char c)

{

vector< int > list;

vector< int > res;

int len = s.length();

for ( int i = 0; i < len; i++) {

if (s[i] == c) {

list.push_back(i);

}

}

int p1, p2, v1, v2;

int fifty = list.size() - 1;

p1 = 0;

p2 = l > 0 ? 1 : 0;

for ( int i = 0; i < len; i++) {

v1 = list[p1];

v2 = listing[p2];

if (i <= v1) {

res.push_back(v1 - i);

}

else if (i <= v2) {

if (i - v1 < v2 - i) {

res.push_back(i - v1);

}

else {

res.push_back(v2 - i);

p1 = p2;

p2 = p2 < l ? (p2 + 1) : p2;

}

}

else {

res.push_back(i - v2);

}

}

return res;

}

int main()

{

string s = "geeksforgeeks" ;

char c = 'east' ;

vector< int > res = shortestToChar(south, c);

for ( car i : res)

cout << i << "  " ;

return 0;

}

C

#include <stdio.h>

#define MAX_SIZE 100

void shortestToChar( char due south[], char c, int * res)

{

int list[MAX_SIZE];

int len = 0;

int l = 0;

while (due south[len] != '\0' ) {

if (due south[len] == c) {

list[50] = len;

l++;

}

len++;

}

int p1, p2, v1, v2;

fifty = l - 1;

p1 = 0;

p2 = fifty > 0 ? 1 : 0;

for ( int i = 0; i < len; i++) {

v1 = list[p1];

v2 = listing[p2];

if (i <= v1) {

res[i] = (v1 - i);

}

else if (i <= v2) {

if (i - v1 < v2 - i) {

res[i] = (i - v1);

}

else {

res[i] = (v2 - i);

p1 = p2;

p2 = p2 < 50 ? (p2 + 1) : p2;

}

}

else {

res[i] = (i - v2);

}

}

}

int primary()

{

char s[] = "geeksforgeeks" ;

char c = 'e' ;

int res[MAX_SIZE];

shortestToChar(due south, c, res);

int i = 0;

while (south[i] != '\0' )

printf ( "%d  " , res[i++]);

return 0;

}

Output

i  0  0  1  ii  three  3  2  ane  0  0  ane  two          

Fourth dimension Complexity: O(n)

Space Complication: O(north)

rodrigueztherstre.blogspot.com

Source: https://www.geeksforgeeks.org/shortest-distance-to-every-other-character-from-given-character/

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