Find Index of 0 to be replaced with 1 to get longest continuous sequence of 1s in a binary array
Last Updated :
24 Jun, 2022
Given an array of 0s and 1s, find the position of 0 to be replaced with 1 to get longest continuous sequence of 1s. Expected time complexity is O(n) and auxiliary space is O(1).
Example:
Input:
arr[] = {1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1}
Output:
Index 9
Assuming array index starts from 0, replacing 0 with 1 at index 9 causes
the maximum continuous sequence of 1s.
Input:
arr[] = {1, 1, 1, 1, 0}
Output:
Index 4
We strongly recommend to minimize the browser and try this yourself first.
A Simple Solution is to traverse the array, for every 0, count the number of 1s on both sides of it. Keep track of maximum count for any 0. Finally return index of the 0 with maximum number of 1s around it. The time complexity of this solution is O(n2).
Using an Efficient Solution, the problem can solved in O(n) time. The idea is to keep track of three indexes, current index (curr), previous zero index (prev_zero) and previous to previous zero index (prev_prev_zero). Traverse the array, if current element is 0, calculate the difference between curr and prev_prev_zero (This difference minus one is the number of 1s around the prev_zero). If the difference between curr and prev_prev_zero is more than maximum so far, then update the maximum. Finally return index of the prev_zero with maximum difference.
Following are the implementations of the above algorithm.
C++
// C++ program to find Index of 0 to be replaced with 1 to get
// longest continuous sequence of 1s in a binary array
#include<iostream>
using namespace std;
// Returns index of 0 to be replaced with 1 to get longest
// continuous sequence of 1s. If there is no 0 in array, then
// it returns -1.
int maxOnesIndex(bool arr[], int n)
{
int max_count = 0; // for maximum number of 1 around a zero
int max_index; // for storing result
int prev_zero = -1; // index of previous zero
int prev_prev_zero = -1; // index of previous to previous zero
// Traverse the input array
for (int curr=0; curr<n; ++curr)
{
// If current element is 0, then calculate the difference
// between curr and prev_prev_zero
if (arr[curr] == 0)
{
// Update result if count of 1s around prev_zero is more
if (curr - prev_prev_zero > max_count)
{
max_count = curr - prev_prev_zero;
max_index = prev_zero;
}
// Update for next iteration
prev_prev_zero = prev_zero;
prev_zero = curr;
}
}
// Check for the last encountered zero
if (n-prev_prev_zero > max_count)
max_index = prev_zero;
return max_index;
}
// Driver program
int main()
{
bool arr[] = {1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1};
int n = sizeof(arr)/sizeof(arr[0]);
cout << "Index of 0 to be replaced is "
<< maxOnesIndex(arr, n);
return 0;
}
// Java program to find Index of 0 to be replaced with 1 to get
// longest continuous sequence of 1s in a binary array
import java.io.*;
class Binary
{
// Returns index of 0 to be replaced with 1 to get longest
// continuous sequence of 1s. If there is no 0 in array, then
// it returns -1.
static int maxOnesIndex(int arr[], int n)
{
int max_count = 0; // for maximum number of 1 around a zero
int max_index=0; // for storing result
int prev_zero = -1; // index of previous zero
int prev_prev_zero = -1; // index of previous to previous zero
// Traverse the input array
for (int curr=0; curr<n; ++curr)
{
// If current element is 0, then calculate the difference
// between curr and prev_prev_zero
if (arr[curr] == 0)
{
// Update result if count of 1s around prev_zero is more
if (curr - prev_prev_zero > max_count)
{
max_count = curr - prev_prev_zero;
max_index = prev_zero;
}
// Update for next iteration
prev_prev_zero = prev_zero;
prev_zero = curr;
}
}
// Check for the last encountered zero
if (n-prev_prev_zero > max_count)
max_index = prev_zero;
return max_index;
}
// Driver program to test above function
public static void main(String[] args)
{
int arr[] = {1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1};
int n = arr.length;
System.out.println("Index of 0 to be replaced is "+
maxOnesIndex(arr, n));
}
}
/* This code is contributed by Devesh Agrawal */
# Python program to find Index
# of 0 to be replaced with 1 to get
# longest continuous sequence
# of 1s in a binary array
# Returns index of 0 to be
# replaced with 1 to get longest
# continuous sequence of 1s.
# If there is no 0 in array, then
# it returns -1.
def maxOnesIndex(arr,n):
# for maximum number of 1 around a zero
max_count = 0
# for storing result
max_index =0
# index of previous zero
prev_zero = -1
# index of previous to previous zero
prev_prev_zero = -1
# Traverse the input array
for curr in range(n):
# If current element is 0,
# then calculate the difference
# between curr and prev_prev_zero
if (arr[curr] == 0):
# Update result if count of
# 1s around prev_zero is more
if (curr - prev_prev_zero > max_count):
max_count = curr - prev_prev_zero
max_index = prev_zero
# Update for next iteration
prev_prev_zero = prev_zero
prev_zero = curr
# Check for the last encountered zero
if (n-prev_prev_zero > max_count):
max_index = prev_zero
return max_index
# Driver program
arr = [1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1]
n = len(arr)
print("Index of 0 to be replaced is ",
maxOnesIndex(arr, n))
# This code is contributed
# by Anant Agarwal.
// C# program to find Index of 0 to be replaced
// with 1 to get longest continuous sequence of
// 1s in a binary array
using System;
class GFG {
// Returns index of 0 to be replaced with
// 1 to get longest continuous sequence of
// 1s. If there is no 0 in array, then it
// returns -1.
static int maxOnesIndex(int []arr, int n)
{
// for maximum number of 1 around a zero
int max_count = 0;
// for storing result
int max_index = 0;
// index of previous zero
int prev_zero = -1;
// index of previous to previous zero
int prev_prev_zero = -1;
// Traverse the input array
for (int curr = 0; curr < n; ++curr)
{
// If current element is 0, then
// calculate the difference
// between curr and prev_prev_zero
if (arr[curr] == 0)
{
// Update result if count of 1s
// around prev_zero is more
if (curr - prev_prev_zero > max_count)
{
max_count = curr - prev_prev_zero;
max_index = prev_zero;
}
// Update for next iteration
prev_prev_zero = prev_zero;
prev_zero = curr;
}
}
// Check for the last encountered zero
if (n-prev_prev_zero > max_count)
max_index = prev_zero;
return max_index;
}
// Driver program to test above function
public static void Main()
{
int []arr = {1, 1, 0, 0, 1, 0, 1, 1, 1,
0, 1, 1, 1};
int n = arr.Length;
Console.Write("Index of 0 to be replaced is "
+ maxOnesIndex(arr, n));
}
}
// This code is contributed by nitin mittal.
<?php
// PHP program to find Index of 0 to be
// replaced with 1 to get longest continuous
// sequence of 1s in a binary array
// Returns index of 0 to be replaced with
// 1 to get longest continuous sequence of 1s.
// If there is no 0 in array, then it returns -1.
function maxOnesIndex( $arr, $n)
{
$max_count = 0; // for maximum number of
// 1 around a zero
$max_index; // for storing result
$prev_zero = -1; // index of previous zero
$prev_prev_zero = -1; // index of previous to
// previous zero
// Traverse the input array
for ($curr = 0; $curr < $n; ++$curr)
{
// If current element is 0, then
// calculate the difference
// between curr and prev_prev_zero
if ($arr[$curr] == 0)
{
// Update result if count of 1s
// around prev_zero is more
if ($curr - $prev_prev_zero > $max_count)
{
$max_count = $curr - $prev_prev_zero;
$max_index = $prev_zero;
}
// Update for next iteration
$prev_prev_zero = $prev_zero;
$prev_zero = $curr;
}
}
// Check for the last encountered zero
if ($n - $prev_prev_zero > $max_count)
$max_index = $prev_zero;
return $max_index;
}
// Driver Code
$arr = array(1, 1, 0, 0, 1, 0, 1,
1, 1, 0, 1, 1, 1);
$n = sizeof($arr);
echo "Index of 0 to be replaced is ",
maxOnesIndex($arr, $n);
// This code is contributed by ajit
?>
<script>
// Javascript program to find Index of 0 to
// be replaced with 1 to get longest continuous
// sequence of 1s in a binary array
// Returns index of 0 to be replaced with
// 1 to get longest continuous sequence of
// 1s. If there is no 0 in array, then it
// returns -1.
function maxOnesIndex(arr, n)
{
// for maximum number of 1 around a zero
let max_count = 0;
// for storing result
let max_index = 0;
// index of previous zero
let prev_zero = -1;
// index of previous to previous zero
let prev_prev_zero = -1;
// Traverse the input array
for(let curr = 0; curr < n; ++curr)
{
// If current element is 0, then
// calculate the difference
// between curr and prev_prev_zero
if (arr[curr] == 0)
{
// Update result if count of 1s
// around prev_zero is more
if (curr - prev_prev_zero > max_count)
{
max_count = curr - prev_prev_zero;
max_index = prev_zero;
}
// Update for next iteration
prev_prev_zero = prev_zero;
prev_zero = curr;
}
}
// Check for the last encountered zero
if (n - prev_prev_zero > max_count)
max_index = prev_zero;
return max_index;
}
// Driver code
let arr = [ 1, 1, 0, 0, 1, 0, 1,
1, 1, 0, 1, 1, 1 ];
let n = arr.length;
document.write("Index of 0 to be replaced is " +
maxOnesIndex(arr, n));
// This code is contributed by divyesh072019
</script>
OutputIndex of 0 to be replaced is 9
Time Complexity: O(n)
Auxiliary Space: O(1)
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