Given a binary array arr[]. The task is to find the position of any 0 in arr[] such that the distance between two set bits is maximized.
Examples
Input: arr = [1, 0, 0, 0, 1, 0, 1]
Output: 2
Explanation: Flip the bit at arr[2]
Input: arr = [1, 0, 0, 0]
Output: 3
Approach: The problem can be solved by finding the longest distance between adjacent set bits with some variation. Follow the steps below to solve the given problem.
- For all distances between adjacent set bits, find the maximum one and store its half as one of the required answers.
- Then find the distance between distance between 0 and the first set bit, and between index N-1 and last set bit.
- Find the overall maximum as the required answer.
- Print the answer found at the end.
Below is the implementation of the above approach.
// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the maximum distance between any
// two set bits after flipping one bit
int maxDistToClosest1(vector<int>& arr)
{
// The size of the array
int n = arr.size(), ans = 0;
int temp = 1, setbit = 0;
// Iterate through the array
for (int i = 1; i < n; i++) {
if (arr[i] == 1) {
if (setbit == 0 && arr[0] == 0)
ans = max(ans, temp);
else
ans = max(ans, temp / 2);
setbit = 1;
temp = 0;
}
temp++;
}
ans = arr[n - 1] == 0 ? max(temp - 1, ans)
: max(temp / 2, ans);
// Return the answer found
return ans;
}
// Driver Code
int main()
{
vector<int> arr = { 1, 0, 0, 0, 1, 0, 1 };
// Function Call
cout << maxDistToClosest1(arr);
return 0;
}
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG {
// Function to find the maximum distance between any
// two set bits after flipping one bit
static int maxDistToClosest1(int arr[])
{
// The size of the array
int n = arr.length, ans = 0;
int temp = 1, setbit = 0;
// Iterate through the array
for (int i = 1; i < n; i++) {
if (arr[i] == 1) {
if (setbit == 0 && arr[0] == 0)
ans = Math.max(ans, temp);
else
ans = Math.max(ans, temp / 2);
setbit = 1;
temp = 0;
}
temp++;
}
ans = arr[n - 1] == 0 ? Math.max(temp - 1, ans)
: Math.max(temp / 2, ans);
// Return the answer found
return ans;
}
// Driver Code
public static void main (String[] args) {
int arr[] = { 1, 0, 0, 0, 1, 0, 1 };
// Function Call
System.out.print(maxDistToClosest1(arr));
}
}
// This code is contributed by hrithikgarg03188.
# Python program for above approach
# Function to find the maximum distance between any
# two set bits after flipping one bit
def maxDistToClosest1(arr):
# The size of the array
n = len(arr)
ans = 0
temp = 1
setbit = 0
# Iterate through the array
for i in range(1, n):
if (arr[i] == 1):
if (setbit == 0 and arr[0] == 0):
ans = max(ans, temp)
else:
ans = max(ans, temp // 2)
setbit = 1
temp = 0
temp +=1
if(arr[n - 1] == 0):
ans = max(temp - 1, ans)
else:
ans = max(temp // 2, ans)
# Return the answer found
return ans
# Driver Code
arr = [ 1, 0, 0, 0, 1, 0, 1 ]
# Function Call
print(maxDistToClosest1(arr))
# This code is contributed by Samim Hossain Mondal.
// C# program for above approach
using System;
class GFG
{
// Function to find the maximum distance between any
// two set bits after flipping one bit
static int maxDistToClosest1(int[] arr)
{
// The size of the array
int n = arr.Length, ans = 0;
int temp = 1, setbit = 0;
// Iterate through the array
for (int i = 1; i < n; i++) {
if (arr[i] == 1) {
if (setbit == 0 && arr[0] == 0)
ans = Math.Max(ans, temp);
else
ans = Math.Max(ans, temp / 2);
setbit = 1;
temp = 0;
}
temp++;
}
ans = arr[n - 1] == 0 ? Math.Max(temp - 1, ans)
: Math.Max(temp / 2, ans);
// Return the answer found
return ans;
}
// Driver Code
public static int Main()
{
int[] arr = { 1, 0, 0, 0, 1, 0, 1 };
// Function Call
Console.Write(maxDistToClosest1(arr));
return 0;
}
}
// This code is contributed by Taranpreet
<script>
// JavaScript program for above approach
// Function to find the maximum distance between any
// two set bits after flipping one bit
const maxDistToClosest1 = (arr) => {
// The size of the array
let n = arr.length, ans = 0;
let temp = 1, setbit = 0;
// Iterate through the array
for (let i = 1; i < n; i++) {
if (arr[i] == 1) {
if (setbit == 0 && arr[0] == 0)
ans = Math.max(ans, temp);
else
ans = Math.max(ans, parseInt(temp / 2));
setbit = 1;
temp = 0;
}
temp++;
}
ans = arr[n - 1] == 0 ? Math.max(temp - 1, ans)
: Math.max(parseInt(temp / 2), ans);
// Return the answer found
return ans;
}
// Driver Code
let arr = [1, 0, 0, 0, 1, 0, 1];
// Function Call
document.write(maxDistToClosest1(arr));
// This code is contributed by rakeshsahni
</script>
Output
2
Time Complexity: O(N)
Auxiliary Space: O(1)
