Given a binary string S, the task is to find the sum of the shortest distance between all 0s to 1 in the given string S.
Examples:
Input: S = “100100”
Output: 5
Explanation:
For the ‘0’ at index 1 the nearest ‘1’ is at index 0 at a distance 1.
For the ‘0’ at index 2 the nearest ‘1’ is at index 3 at a distance 1.
For the ‘0’ at index 4 the nearest ‘1’ is at index 3 at a distance 1.
For the ‘0’ at index 5 the nearest ‘1’ is at index 3 at a distance 2.
Therefore the sum of the distances is 1 + 1 + 1 + 2 = 5.Input: S = “1111”
Output: 0
Approach: The given problem can be solved by using the Greedy Approach. The idea is to search for the ‘1′ for each ‘0′ which is nearest to it from the left side. Similarly, search for the ‘1′ for each ‘0’ which is nearest to it from the right side. Finally, calculate the sum of the distances which is minimum for any ‘0′ to ‘1′ from the calculated value of right and left sides. Follow the below steps to solve the given problem.
- Initialize the two arrays prefixDistance(N), suffixDistance(N) to store distance from left and right respectively.
- Initialize a variable, say cnt = 0 that store the distance between any ‘0′ to nearest ‘1′.
- Initialize a variable, say haveOne = false, to mark the character ‘1′.
- Initialize a variable, say sum = 0 that stores the total sum between all the ‘0′ to its nearest ‘1′.
-
Iterate over the range [0, N – 1] using the variable i perform the following steps:
- If the value of S[i] is ‘1′ then assign haveOne as true, cnt as 0 and prefixDistance[i] as 0.
- Otherwise, if haveOne is true then increment the value of cnt by 1 and assign the value of prefixDistance[i] as cnt.
- Iterate over the range [0, N – 1] using the variable i perform the following steps:
- If the value of S[i] is ‘1′ then assign haveOne as true, cnt as 0 and suffixDistance[i] as 0.
- Otherwise, if haveOne is true then increment the value of cnt by 1 and assign the value of suffixDistance[i] as cnt.
- Iterate over the range [0, N – 1] using the variable i and if the value of S[i] is ‘1′ then update the sum as sum += min(prefixDistance[i], suffixDistance[i]).
- After completing the above steps, print the value of the sum as the result.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the total sum of // the shortest distance between every // 0 to 1 in a given binary string void findTotalDistance(string S, int N)
{ // Stores the prefix distance and
// suffix distance from 0 to 1
vector< int > prefixDistance(N);
vector< int > suffixDistance(N);
// Stores the current distance
// from 1 to 0
int cnt = 0;
// Marks the 1
bool haveOne = false ;
for ( int i = 0; i < N; ++i) {
// If current character is 1
if (S[i] == '1' ) {
// Mark haveOne to true
haveOne = true ;
// Assign the cnt to 0
cnt = 0;
// Assign prefixDistance[i] as 0
prefixDistance[i] = 0;
}
// If haveOne is true
else if (haveOne) {
// Update the cnt
cnt++;
// Update prefixDistance[i]
prefixDistance[i] = cnt;
}
// Assign prefixDistance[i]
// as INT_MAX
else
prefixDistance[i] = INT_MAX;
}
// Assign haveOne as false
haveOne = false ;
for ( int i = N - 1; i >= 0; --i) {
// If current character is 1
if (S[i] == '1' ) {
// Mark haveOne to true
haveOne = true ;
// Assign the cnt to 0
cnt = 0;
// Assign the suffixDistance[i]
// as 0
suffixDistance[i] = 0;
}
// If haveOne is true
else if (haveOne) {
// Update the cnt
cnt++;
// Update suffixDistance[i]
// as cnt
suffixDistance[i] = cnt;
}
else
// Assign suffixDistance[i]
// as INT_MAX
suffixDistance[i] = INT_MAX;
}
// Stores the total sum of distances
// between 0 to nearest 1
int sum = 0;
for ( int i = 0; i < N; ++i) {
// If current character is 0
if (S[i] == '0' ) {
// Update the value of sum
sum += min(prefixDistance[i],
suffixDistance[i]);
}
}
// Print the value of the sum
cout << sum << endl;
} // Driver Code int main()
{ string S = "100100" ;
int N = S.length();
findTotalDistance(S, N);
return 0;
} |
// Java program for the above approach import java.io.*;
public class GFG{
// Function to find the total sum of // the shortest distance between every // 0 to 1 in a given binary string static void findTotalDistance(String S, int N)
{ // Stores the prefix distance and
// suffix distance from 0 to 1
int []prefixDistance = new int [N];
int []suffixDistance = new int [N];
// Stores the current distance
// from 1 to 0
int cnt = 0 ;
// Marks the 1
boolean haveOne = false ;
for ( int i = 0 ; i < N; ++i) {
// If current character is 1
if (S.charAt(i) == '1' ) {
// Mark haveOne to true
haveOne = true ;
// Assign the cnt to 0
cnt = 0 ;
// Assign prefixDistance[i] as 0
prefixDistance[i] = 0 ;
}
// If haveOne is true
else if (haveOne) {
// Update the cnt
cnt++;
// Update prefixDistance[i]
prefixDistance[i] = cnt;
}
// Assign prefixDistance[i]
// as INT_MAX
else
prefixDistance[i] = Integer.MAX_VALUE;
}
// Assign haveOne as false
haveOne = false ;
for ( int i = N - 1 ; i >= 0 ; --i) {
// If current character is 1
if (S.charAt(i) == '1' ) {
// Mark haveOne to true
haveOne = true ;
// Assign the cnt to 0
cnt = 0 ;
// Assign the suffixDistance[i]
// as 0
suffixDistance[i] = 0 ;
}
// If haveOne is true
else if (haveOne) {
// Update the cnt
cnt++;
// Update suffixDistance[i]
// as cnt
suffixDistance[i] = cnt;
}
else
// Assign suffixDistance[i]
// as INT_MAX
suffixDistance[i] = Integer.MAX_VALUE;
}
// Stores the total sum of distances
// between 0 to nearest 1
int sum = 0 ;
for ( int i = 0 ; i < N; ++i) {
// If current character is 0
if (S.charAt(i) == '0' ) {
// Update the value of sum
sum += Math.min(prefixDistance[i],
suffixDistance[i]);
}
}
// Print the value of the sum
System.out.print(sum);
} // Driver Code public static void main(String []args)
{ String S = "100100" ;
int N = S.length();
findTotalDistance(S, N);
} } // This code is contributed by AnkThon |
# python program for the above approach # Function to find the total sum of # the shortest distance between every # 0 to 1 in a given binary string INT_MAX = 2147483647
def findTotalDistance(S, N):
# Stores the prefix distance and
# suffix distance from 0 to 1
prefixDistance = [ 0 for _ in range (N)]
suffixDistance = [ 0 for _ in range (N)]
# Stores the current distance
# from 1 to 0
cnt = 0
# Marks the 1
haveOne = False
for i in range ( 0 , N):
# If current character is 1
if (S[i] = = '1' ):
# // Mark haveOne to true
haveOne = True
# Assign the cnt to 0
cnt = 0
# Assign prefixDistance[i] as 0
prefixDistance[i] = 0
# If haveOne is true
elif (haveOne):
# Update the cnt
cnt = cnt + 1
# Update prefixDistance[i]
prefixDistance[i] = cnt
# Assign prefixDistance[i]
# as INT_MAX
else :
prefixDistance[i] = INT_MAX
# Assign haveOne as false
haveOne = False
for i in range (N - 1 , - 1 , - 1 ):
# If current character is 1
if (S[i] = = '1' ):
# Mark haveOne to true
haveOne = True
# Assign the cnt to 0
cnt = 0
# Assign the suffixDistance[i]
# as 0
suffixDistance[i] = 0
# If haveOne is true
elif (haveOne):
# Update the cnt
cnt = cnt + 1
# Update suffixDistance[i]
# as cnt
suffixDistance[i] = cnt
else :
# // Assign suffixDistance[i]
# // as INT_MAX
suffixDistance[i] = INT_MAX
# Stores the total sum of distances
# between 0 to nearest 1
sum = 0
for i in range ( 0 , N):
# If current character is 0
if (S[i] = = '0' ):
# Update the value of sum
sum + = min (prefixDistance[i], suffixDistance[i])
# Print the value of the sum
print ( sum )
# Driver Code if __name__ = = "__main__" :
S = "100100"
N = len (S)
findTotalDistance(S, N)
# This code is contributed by rakeshsahni |
// C# program for the above approach using System;
using System.Collections.Generic;
class GFG{
// Function to find the total sum of // the shortest distance between every // 0 to 1 in a given binary string static void findTotalDistance( string S, int N)
{ // Stores the prefix distance and
// suffix distance from 0 to 1
int []prefixDistance = new int [N];
int []suffixDistance = new int [N];
// Stores the current distance
// from 1 to 0
int cnt = 0;
// Marks the 1
bool haveOne = false ;
for ( int i = 0; i < N; ++i) {
// If current character is 1
if (S[i] == '1' ) {
// Mark haveOne to true
haveOne = true ;
// Assign the cnt to 0
cnt = 0;
// Assign prefixDistance[i] as 0
prefixDistance[i] = 0;
}
// If haveOne is true
else if (haveOne) {
// Update the cnt
cnt++;
// Update prefixDistance[i]
prefixDistance[i] = cnt;
}
// Assign prefixDistance[i]
// as INT_MAX
else
prefixDistance[i] = Int32.MaxValue;
}
// Assign haveOne as false
haveOne = false ;
for ( int i = N - 1; i >= 0; --i) {
// If current character is 1
if (S[i] == '1' ) {
// Mark haveOne to true
haveOne = true ;
// Assign the cnt to 0
cnt = 0;
// Assign the suffixDistance[i]
// as 0
suffixDistance[i] = 0;
}
// If haveOne is true
else if (haveOne) {
// Update the cnt
cnt++;
// Update suffixDistance[i]
// as cnt
suffixDistance[i] = cnt;
}
else
// Assign suffixDistance[i]
// as INT_MAX
suffixDistance[i] = Int32.MaxValue;
}
// Stores the total sum of distances
// between 0 to nearest 1
int sum = 0;
for ( int i = 0; i < N; ++i) {
// If current character is 0
if (S[i] == '0' ) {
// Update the value of sum
sum += Math.Min(prefixDistance[i],
suffixDistance[i]);
}
}
// Print the value of the sum
Console.Write(sum);
} // Driver Code public static void Main()
{ string S = "100100" ;
int N = S.Length;
findTotalDistance(S, N);
} } // This code is contributed by SURENDRA_GANGWAR. |
<script> // JavaScript program for the above approach
const INT_MAX = 2147483647
// Function to find the total sum of
// the shortest distance between every
// 0 to 1 in a given binary string
const findTotalDistance = (S, N) => {
// Stores the prefix distance and
// suffix distance from 0 to 1
let prefixDistance = new Array(N).fill(0);
let suffixDistance = new Array(N).fill(0);
// Stores the current distance
// from 1 to 0
let cnt = 0;
// Marks the 1
let haveOne = false ;
for (let i = 0; i < N; ++i) {
// If current character is 1
if (S[i] == '1' ) {
// Mark haveOne to true
haveOne = true ;
// Assign the cnt to 0
cnt = 0;
// Assign prefixDistance[i] as 0
prefixDistance[i] = 0;
}
// If haveOne is true
else if (haveOne) {
// Update the cnt
cnt++;
// Update prefixDistance[i]
prefixDistance[i] = cnt;
}
// Assign prefixDistance[i]
// as INT_MAX
else
prefixDistance[i] = INT_MAX;
}
// Assign haveOne as false
haveOne = false ;
for (let i = N - 1; i >= 0; --i) {
// If current character is 1
if (S[i] == '1' ) {
// Mark haveOne to true
haveOne = true ;
// Assign the cnt to 0
cnt = 0;
// Assign the suffixDistance[i]
// as 0
suffixDistance[i] = 0;
}
// If haveOne is true
else if (haveOne) {
// Update the cnt
cnt++;
// Update suffixDistance[i]
// as cnt
suffixDistance[i] = cnt;
}
else
// Assign suffixDistance[i]
// as INT_MAX
suffixDistance[i] = INT_MAX;
}
// Stores the total sum of distances
// between 0 to nearest 1
let sum = 0;
for (let i = 0; i < N; ++i) {
// If current character is 0
if (S[i] == '0' ) {
// Update the value of sum
sum += Math.min(prefixDistance[i], suffixDistance[i]);
}
}
// Print the value of the sum
document.write(`${sum}<br/>`);
}
// Driver Code
let S = "100100" ;
let N = S.length;
findTotalDistance(S, N);
// This code is contributed by rakeshsahni </script> |
5
Time Complexity: O(N)
Auxiliary Space: O(N)