PROBLEM DESCRIPTION
Given a binary string A. It is allowed to do at most one swap between any 0 and 1. Find and return the length of the longest consecutive 1’s that can be achieved.
SOLUTION
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public class Solution {
public int solve(String s) {
int n = s.length();
//convert string to integer array
int[] A = new int[n];
for(int i=0; i<n; i++){
A[i] = s.charAt(i)=='0'?0:1;
}
//number of consecutive 1s on the left of the element
int[] left = new int[n];
//number of consecutive 1s on the right of the element
int[] right = new int[n];
//there is no element on the left of first
left[0] = 0;
//there is no element on the right of last
right[n-1] = 0;
//calculate number of consecutive 1s for the other elements on the left of it
for(int i=1; i<n; i++){
//if the previous element was 1, add 1
if(A[i-1] == 1){
left[i] = left[i-1] + 1;
//else reset to 0
}else{
left[i] = 0;
}
}
//calculate number of consecutive 1s for the other elements on the right of it
for(int i=n-2; i>=0; i--){
//if the previous element was 1, add 1
if(A[i+1] == 1){
right[i] = right[i+1] + 1;
//else reset to 0
}else{
right[i] = 0;
}
}
//total number of 1s present in the given array
int countOfOne = 0;
for(int i=0; i<n; i++) if(A[i] == 1) countOfOne++;
//if the number of 1s is equal to the array size, we can return N
if(countOfOne == n) return n; //edge case
//init: min possible answer is 0 if there are no 1s
int maxLength = 0;
//for each element, check what is the max length with consecutive 1s possible and keep updating maxLength
for(int i=0; i<n; i++){
//if the current element is 0, check if it can be replaced with 1
if(A[i] == 0){
//number of 1s on its left
int l = left[i];
//number of 1s on its right
int r = right[i];
//we can replace it with 1 only if there are any extra 1s remaining
//to find that, add the number of 1s on left and right and check if the sum if less than the total number of 1s in the array
//if it's lesser, there must be an extra 1 somewhere which can be swapped
boolean anyOneRemaining = (countOfOne-(l+r))>0;
//if there is an extra 1 which can be swapped, the total length possible would be: L+R+1
if(anyOneRemaining){
maxLength = Math.max(maxLength, l+r+1);
//if extra 1 is not present, then the best possible option is to take the left most or right most 1 and swap it with 0
//in that case, the length will be l+r
//Example: 1110111 -> here we can take the left most or right most 1 and swap with 0
}else{
maxLength = Math.max(maxLength, l+r);
}
}
}
return maxLength;
}
}