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++
// 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 bitint 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 Codeint main(){ vector<int> arr = { 1, 0, 0, 0, 1, 0, 1 }; // Function Call cout << maxDistToClosest1(arr); return 0;} |
Java
// Java program for the above approachimport 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
# Python program for above approach# Function to find the maximum distance between any# two set bits after flipping one bitdef 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 Codearr = [ 1, 0, 0, 0, 1, 0, 1 ]# Function Callprint(maxDistToClosest1(arr))# This code is contributed by Samim Hossain Mondal. |
C#
// C# program for above approachusing 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 |
Javascript
<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)
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