Queries to answer the number of ones and zero to the left of given index

Given a binary array and Q queries. Every query consists of a number K, the task is to print the number of ones and zeros to the left of the index K. 
Examples:

Input: arr[] = {1, 1, 1, 0, 0, 1, 0, 1, 1}, Q[] = {0, 1, 2, 4} 
Output:
0 ones 0 zeros 
1 ones 0 zeros 
2 ones 0 zeros 
3 ones 1 zeros
Input: arr[] = {1, 0, 1, 0, 1, 1}, Q[] = {2, 3} 
Output:
1 ones 1 zeros 
2 ones 1 zeros 

Approach: The following steps can be followed to solve the above problem: 

• Declare an array of pairs(say left[]) which will be used to pre-calculate the number of ones and zeros to the left.
• Iterate on the array and in every step initialize left[i] with the number of ones and zeros to the left.
• The number of ones for every query will be left[k].first and number of zeros will be left[k].second

Below is the implementation of the above approach: 

C++

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

// Function to pre-calculate the left[] array
void preCalculate(int binary[], int n, pair<int, int> left[])
{
    int count1 = 0, count0 = 0;

    // Iterate in the binary array
    for (int i = 0; i < n; i++) {

        // Initialize the number
        // of 1 and 0
        left[i].first = count1;
        left[i].second = count0;

        // Increase the count
        if (binary[i])
            count1++;
        else
            count0++;
    }
}

// Driver code
int main()
{

    int binary[] = { 1, 1, 1, 0, 0, 1, 0, 1, 1 };
    int n = sizeof(binary) / sizeof(binary[0]);
    pair<int, int> left[n];
    preCalculate(binary, n, left);

    // Queries
    int queries[] = { 0, 1, 2, 4 };
    int q = sizeof(queries) / sizeof(queries[0]);

    // Solve queries
    for (int i = 0; i < q; i++)
        cout << left[queries[i]].first << " ones "
             << left[queries[i]].second << " zeros\n";

    return 0;
}

Java

// Java implementation of the approach
class GFG {

    // pair class
    static class pair {
        int first, second;
        pair(int a, int b)
        {
            first = a;
            second = b;
        }
    }

    // Function to pre-calculate the left[] array
    static void preCalculate(int binary[], int n,
                             pair left[])
    {
        int count1 = 0, count0 = 0;

        // Iterate in the binary array
        for (int i = 0; i < n; i++) {

            // Initialize the number
            // of 1 and 0
            left[i].first = count1;
            left[i].second = count0;

            // Increase the count
            if (binary[i] != 0)
                count1++;
            else
                count0++;
        }
    }

    // Driver code
    public static void main(String args[])
    {

        int binary[] = { 1, 1, 1, 0, 0, 1, 0, 1, 1 };
        int n = binary.length;
        pair left[] = new pair[n];

        for (int i = 0; i < n; i++)
            left[i] = new pair(0, 0);

        preCalculate(binary, n, left);

        // Queries
        int queries[] = { 0, 1, 2, 4 };
        int q = queries.length;

        // Solve queries
        for (int i = 0; i < q; i++)
            System.out.println(left[queries[i]].first + " ones "
                               + left[queries[i]].second + " zeros\n");
    }
}

// This code is contributed by Arnab Kundu

Python3

# Python3 implementation of the approach

# Function to pre-calculate the left[] array
def preCalculate(binary, n, left):

    count1, count0 = 0, 0

    # Iterate in the binary array
    for i in range(n):

        # Initialize the number
        # of 1 and 0
        left[i][0] = count1
        left[i][1] = count0

        # Increase the count
        if (binary[i]):
            count1 += 1
        else:
            count0 += 1

# Driver code
binary = [1, 1, 1, 0, 0, 1, 0, 1, 1]

n = len(binary)

left = [[ 0 for i in range(2)]
            for i in range(n)]

preCalculate(binary, n, left)

queries = [0, 1, 2, 4 ]

q = len(queries)

# Solve queries
for i in range(q):
    print(left[queries[i]][0], "ones",
          left[queries[i]][1], "zeros")

# This code is contributed
# by mohit kumar

C#

// C# implementation of the approach
using System;

class GFG {

    // pair class
    public class pair {
        public int first, second;
        public pair(int a, int b)
        {
            first = a;
            second = b;
        }
    }

    // Function to pre-calculate the left[] array
    static void preCalculate(int[] binary, int n,
                             pair[] left)
    {
        int count1 = 0, count0 = 0;

        // Iterate in the binary array
        for (int i = 0; i < n; i++) {

            // Initialize the number
            // of 1 and 0
            left[i].first = count1;
            left[i].second = count0;

            // Increase the count
            if (binary[i] != 0)
                count1++;
            else
                count0++;
        }
    }

    // Driver code
    public static void Main(String[] args)
    {

        int[] binary = { 1, 1, 1, 0, 0, 1, 0, 1, 1 };
        int n = binary.Length;
        pair[] left = new pair[n];

        for (int i = 0; i < n; i++)
            left[i] = new pair(0, 0);

        preCalculate(binary, n, left);

        // Queries
        int[] queries = { 0, 1, 2, 4 };
        int q = queries.Length;

        // Solve queries
        for (int i = 0; i < q; i++)
            Console.WriteLine(left[queries[i]].first + " ones "
                              + left[queries[i]].second + " zeros\n");
    }
}

// This code contributed by Rajput-Ji

PHP

<?php
// PHP implementation of the approach

// Function to pre-calculate the left[] array
function preCalculate($binary, $n)
{
    $left = array();
    $count1 = 0; $count0 = 0;

    // Iterate in the binary array
    for ($i = 0; $i < $n; $i++)
    {

        // Initialize the number
        // of 1 and 0
        $left[$i] = array($count1,
                          $count0);

        // Increase the count
        if ($binary[$i])
            $count1++;
        else
            $count0++;
    }
    return $left;
}

// Driver code
$binary = array( 1, 1, 1, 0, 0, 1, 0, 1, 1 );
$n = count($binary);

$left = preCalculate($binary, $n);

// Queries
$queries = array( 0, 1, 2, 4 );
$q = count($queries);

// Solve queries
for ($i = 0; $i < $q; $i++)
    echo $left[$queries[$i]][0], " ones ",
         $left[$queries[$i]][1], " zeros\n";

// This code is contributed by Ryuga
?>

Javascript

<script>
// Javascript implementation of the approach

// Function to pre-calculate the left[] array
function preCalculate(binary,n,left)
{
    let count1 = 0, count0 = 0;

        // Iterate in the binary array
        for (let i = 0; i < n; i++) {

            // Initialize the number
            // of 1 and 0
            left[i][0] = count1;
            left[i][1] = count0;

            // Increase the count
            if (binary[i] != 0)
                count1++;
            else
                count0++;
        }
}

// Driver code
let binary=[1, 1, 1, 0, 0, 1, 0, 1, 1];
let n = binary.length;

let left=new Array(n);

    for (let i = 0; i < n; i++)
            left[i] = [0, 0];

        preCalculate(binary, n, left);

        // Queries
        let queries = [ 0, 1, 2, 4 ];
        let q = queries.length;

        // Solve queries
        for (let i = 0; i < q; i++)
            document.write(left[queries[i]][0] + " ones "
                               + left[queries[i]][1] + " zeros<br>");


// This code is contributed by unknown2108
</script>

Output:

0 ones 0 zeros
1 ones 0 zeros
2 ones 0 zeros
3 ones 1 zeros

Time Complexity: O(N+q), as for pre-computation will take O(N) time because we are using a loop to traverse N times and O(1) for every query so total time will be O(q).
Auxiliary Space: O(N), as we are using extra space for the array left.