Check if all elements of binary array can be made 1

Last Updated : 25 Sep, 2022

Given a binary array Arr and an integer K. If the value at index i is 1 you can change 0 to 1 with in the range of ( i - K ) to ( i + K ).
The task is to determine whether all the elements of the array can be made 1 or not.
Examples:

Input: Arr = { 0, 1, 0, 1 }, K = 2
Output: 2

Input: Arr = { 1, 0, 0, 0, 0, 0, 1 }, K = 2
Output: 0
It is not possible to make all the elements equal to 1

Approach:
Here another array is being used to mark as 1 if we can reach that index.
For every index in the range of 1 to N if the value of Arr[i] is 1 then make a loop from (i - K) to (i + K) and update b[i] to 1.
At last check the entry of b[i], and it should be 1 for every i, if it is then print 1 else print 0.
Below is the implementation of the above approach:

C++

// C++ implementation
#include <bits/stdc++.h>
using namespace std;

// Function to print 1 if the
// it is possible to make all array
// element equal to 1 else 0
void checkAllOnes(int arr[], int n,
                 int k)
{

    int brr[n];

    // Iterating over the array
    for (int i = 0; i < n; i++) {

        // If element is 1
        if (arr[i] == 1) {

            int h = k + 1;
            int j = i;

            // Put b[j...j-k] = 0
            while (j >= 0 && (h--)) {
                brr[j] = 1;
                j--;
            }

            h = k + 1;
            j = i;

            // Put b[j...j+k] = 0
            while (j < n && (h--)) {
                brr[j] = 1;
                j++;
            }
        }
    }

    int flag = 0;

    // If any value in aux
    // array equal to 0
    // then set flag
    for (int i = 0; i < n; i++) {
        if (brr[i] == 0) {
            flag = 1;
            break;
        }
    }

    // If flag is set this
    // means array after
    // conversion contains
    // 0 so print 0
    if (flag == 1)
        cout << "0";

    // else print 1
    else
        cout << "1\n";
}

// Driver Code
int main()
{

    int arr[] = { 1, 0, 1, 0 };
    int k = 2;
    int n = sizeof(arr) / sizeof(arr[0]);

    checkAllOnes(arr, n, k);

    return 0;
}

Java

// Java implementation of above approach
class GFG 
{
        
// Function to print 1 if the 
// it is possible to make all array 
// element equal to 1 else 0 
static void checkAllOnes(int arr[], 
                         int n, int k) 
{ 
    int brr[] = new int[n]; 

    // Iterating over the array 
    for (int i = 0; i < n; i++) 
    { 

        // If element is 1 
        if (arr[i] == 1) 
        { 
            int h = k + 1; 
            int j = i; 

            // Put b[j...j-k] = 0 
            while ((j >= 0) && (h-- != 0))
            { 
                brr[j] = 1; 
                j--; 
            } 

            h = k + 1; 
            j = i; 

            // Put b[j...j+k] = 0 
            while ((j < n) && (h-- != 0)) 
            { 
                brr[j] = 1; 
                j++; 
            } 
        } 
    } 

    int flag = 0; 

    // If any value in aux 
    // array equal to 0 
    // then set flag 
    for (int i = 0; i < n; i++) 
    { 
        if (brr[i] == 0) 
        { 
            flag = 1; 
            break; 
        } 
    } 

    // If flag is set this 
    // means array after 
    // conversion contains 
    // 0 so print 0 
    if (flag == 1) 
        System.out.println("0"); 

    // else print 1 
    else
        System.out.println("1"); 
} 

// Driver Code 
public static void main (String[] args) 
{ 
    int arr[] = { 1, 0, 1, 0 }; 
    int k = 2; 
    int n = arr.length; 

    checkAllOnes(arr, n, k); 
} 
}

// This code is contributed by AnkitRai01

Python3

# Python3 implementation

# Function to print 1 if the
# it is possible to make all array
# element equal to 1 else 0
def checkAllOnes(arr, n, k):

    brr = [0 for i in range(n)]

    # Iterating over the array
    for i in range(n):

        # If element is 1
        if (arr[i] == 1):

            h = k + 1
            j = i

            # Put b[j...j-k] = 0
            while (j >= 0 and (h)):
                brr[j] = 1
                h -= 1
                j -= 1

            h = k + 1
            j = i

            # Put b[j...j+k] = 0
            while (j < n and (h)):
                brr[j] = 1
                j += 1
                h -= 1

    flag = 0

    # If any value in aux
    # array equal to 0
    # then set flag
    for i in range(n):
        if (brr[i] == 0):
            flag = 1
            break

    # If flag is set this
    # means array after
    # conversion contains
    # 0 so pr0
    if (flag == 1):
        print("0")

    # else pr1
    else:
        print("1")

# Driver Code
arr = [1, 0, 1, 0]
k = 2
n = len(arr)

checkAllOnes(arr, n, k)

# This code is contributed by Mohit Kumar

// C# implementation of above approach
using System;
                    
class GFG 
{
        
// Function to print 1 if the 
// it is possible to make all array 
// element equal to 1 else 0 
static void checkAllOnes(int []arr, 
                         int n, int k) 
{ 
    int []brr = new int[n]; 

    // Iterating over the array 
    for (int i = 0; i < n; i++) 
    { 

        // If element is 1 
        if (arr[i] == 1) 
        { 
            int h = k + 1; 
            int j = i; 

            // Put b[j...j-k] = 0 
            while ((j >= 0) && (h-- != 0))
            { 
                brr[j] = 1; 
                j--; 
            } 

            h = k + 1; 
            j = i; 

            // Put b[j...j+k] = 0 
            while ((j < n) && (h-- != 0)) 
            { 
                brr[j] = 1; 
                j++; 
            } 
        } 
    } 

    int flag = 0; 

    // If any value in aux 
    // array equal to 0 
    // then set flag 
    for (int i = 0; i < n; i++) 
    { 
        if (brr[i] == 0) 
        { 
            flag = 1; 
            break; 
        } 
    } 

    // If flag is set this 
    // means array after 
    // conversion contains 
    // 0 so print 0 
    if (flag == 1) 
        Console.WriteLine("0"); 

    // else print 1 
    else
        Console.WriteLine("1"); 
} 

// Driver Code 
public static void Main (String[] args) 
{ 
    int []arr = { 1, 0, 1, 0 }; 
    int k = 2; 
    int n = arr.Length; 

    checkAllOnes(arr, n, k); 
} 
}

// This code is contributed by 29AjayKumar

JavaScript

<script>

// Javascript implementation

// Function to print 1 if the
// it is possible to make all array
// element equal to 1 else 0
function checkAllOnes(arr, n, k)
{

    let brr = new Array(n);

    // Iterating over the array
    for (let i = 0; i < n; i++) {

        // If element is 1
        if (arr[i] == 1) {

            let h = k + 1;
            let j = i;

            // Put b[j...j-k] = 0
            while (j >= 0 && (h--)) {
                brr[j] = 1;
                j--;
            }

            h = k + 1;
            j = i;

            // Put b[j...j+k] = 0
            while (j < n && (h--)) {
                brr[j] = 1;
                j++;
            }
        }
    }

    let flag = 0;

    // If any value in aux
    // array equal to 0
    // then set flag
    for (let i = 0; i < n; i++) {
        if (brr[i] == 0) {
            flag = 1;
            break;
        }
    }

    // If flag is set this
    // means array after
    // conversion contains
    // 0 so print 0
    if (flag == 1)
        document.write("0");

    // else print 1
    else
        document.write("1");
}

// Driver Code

    let arr = [ 1, 0, 1, 0 ];
    let k = 2;
    let n = arr.length;

    checkAllOnes(arr, n, k);

</script>

Output:

Time complexity: O(n) where n is the number of elements in the given array.
Auxiliary space: O(n), as using extra space for array brr.

Check if Array elements can be made 0 by Subtracting left adjacent values

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Check if all elements of binary array can be made 1

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