Replace the middle element of the longest subarray of 0s from the right exactly K times

Difficulty Level : Easy
Last Updated : 04 Jan, 2021

Given an array arr[] of size N, consisting of 0s initially, and a positive integer K, the task is to print the array elements by performing the following operations exactly K times.

For every i^th operation select the rightmost longest subarray consisting of all 0s and replace the mid element of the subarray by i.
If two middle elements exist, then check if i is an even number or not. If found to be true, then replace the rightmost middle element with i.
Otherwise, replace the leftmost middle element with i.

Examples:

Input: arr[] = { 0, 0, 0, 0, 0}, K = 3
Output: 3 0 1 0 2
Explanation:
In 1st operation selecting the subarray { arr[0], …, arr[4]} and replacing arr[2] by 1 modifies arr[] to { 0, 0, 1, 0, 0}
In 2nd operation selecting the subarray { arr[3], …, arr[4]} and replacing arr[4] by 2 modifies arr[] to { 0, 0, 1, 0, 2}
In 3rd operation selecting the subarray { arr[0], …, arr[1]} and replacing arr[1] by 3 modifies arr[] to { 0, 3, 1, 0, 2}
Therefore, the required output is 3 0 1 0 2.

Input: arr[] = { 0, 0, 0, 0, 0, 0, 0 }, K = 7
Output: 7 3 6 1 5 2 4

Approach: The problem can be solved using Greedy technique. The idea is to use the Priority Queue to select the rightmost longest subarray with all 0s. Follow the steps below to solve the problem:

Initialize a Priority Queue, say pq, to store the subarrays of the form { X, Y} where X denotes the length of the subarray and Y denotes the starting index of the subarray.
Initially maximum length of the subarray with all 0s is N and start index of the subarray is 0. Therefore, Insert { N, 0 } into pq.
Iterate over the range [1, K] using variable i. For every i^th operation pop the top element from pq and check if length of the popped element is an odd number or not. If found to be true then replace the mid element of the subarray with i.
Otherwise, if i is an even number then replace the rightmost mid element of the subarray with i. Otherwise, replace the leftmost mid element of the subarray with i.
After replacing the mid element with i, insert the left half of the subarray and right half of the subarray containing all 0s into pq.
Finally, print the array elements.

Below is the implementation of the above approach:

C++

// C++ program to implement 
// the above approach 
  
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to print array by replacing the mid 
// of the righmost longest subarray with count 
// of operations performed on the array  
void ReplaceArray(int arr[], int N, int K) 
{ 
      
    // Stores subarray of the form { X, Y }, 
    // where X is the length and Y is start 
    // index of the subarray 
    priority_queue<vector<int> > pq; 
      
    // Insert the array arr[]  
    pq.push({ N, 0 }); 
      
    // Stores index of mid  
    // element of the subarray 
    int mid; 
      
    // Iterate over the range [1, N] 
    for (int i = 1; i <= K; i++) { 
          
        // Stores top element of pq 
        vector<int> sub = pq.top(); 
          
        // Pop top element of pq 
        pq.pop(); 
          
        // If length of the subarray 
        // is an odd number 
        if (sub[0] % 2 == 1) { 
              
            // Update mid 
            mid = sub[1] + sub[0] / 2; 
              
            // Replacing arr[mid] with i 
            arr[mid] = i; 
              
            // Insert left half of 
            // the subarray into pq 
            pq.push({ sub[0] / 2, 
                         sub[1] }); 
              
            // Insert right half of 
            // the subarray into pq 
            pq.push({ sub[0] / 2,  
                           (mid + 1) }); 
        } 
          
        // If length of the current  
        // subarray is an even number 
        else { 
              
            // If i is  
            // an odd number 
            if (i % 2 == 1) { 
                  
                // Update mid 
                mid = sub[1] + sub[0] / 2; 
                  
                // Replacing mid element 
                // with i 
                arr[mid - 1] = i; 
                  
                // Insert left half of 
                // the subarray into pq 
                pq.push({ sub[0] / 2 - 1,  
                               sub[1] }); 
                  
                // Insert right half of 
                // the subarray into pq 
                pq.push({ sub[0] / 2, mid }); 
            } 
              
            // If i is an even number 
            else { 
                  
                // Update mid 
                mid = sub[1] + sub[0] / 2; 
                  
                // Replacing mid element 
                // with i 
                arr[mid - 1] = i; 
                  
                // Insert left half of 
                // the subarray into pq 
                pq.push({ sub[0] / 2, 
                            sub[1] }); 
                  
                // Insert right half of 
                // the subarray into pq 
                pq.push({ sub[0] / 2 - 1, 
                           (mid + 1) }); 
            } 
        } 
    } 
      
    // Print array elements 
    for (int i = 0; i < N; i++) 
        cout << arr[i] << " "; 
} 
  
// Driver Code 
int main() 
{ 
      
    int arr[] = { 0, 0, 0, 0, 0 }; 
    int N = sizeof(arr) / sizeof(arr[0]); 
    int K = 3; 
    ReplaceArray(arr, N, K); 
}

Output:

3 0 1 2 0

Time Complexity: O(K * log(N))
Auxiliary Space: O(N)

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