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Make all the elements of array odd by incrementing odd-indexed elements of odd-length subarrays
  • Difficulty Level : Medium
  • Last Updated : 20 Apr, 2021

Given an array arr[] of size N, the task is to make all the array elements odd by choosing an odd length subarray of arr[] and increment all odd positioned elements by 1 in this subarray. Print the count of such operations required.

Examples:

Input: arr[] = {2, 3, 4, 3, 5, 3, 2}
Output: 2
Explanation:
In first operation, choose the subarray {2, 3, 4} and increment all its elements at odd positions, i.e., 1 and 3 of this subarray. The updated array is {3, 3, 5, 3, 5, 3, 2}.
In second operation, choose the last index which is subarray of length 1 and increment its value. The updated array is {3, 3, 5, 3, 5, 3, 3}

Input: arr[] = {1, 5, 7}
Output: 0
Explanation: Since all array elements are odd, no changes required.

Approach: The idea is based on the observation that whenever a subarray is chosen, either the odd positioned values are changed or the even positioned values in the original array. The problem can be solved by choosing the subarray greedily in each operation. First, iterate over all the odd indices and mark the starting of the subarray as soon as an even value is found, and end the subarray when an odd value is found, simultaneously updating the number of operations. Repeat the same process for even indices. Follow the steps below to solve the problem:



  1. Initialize a variable, say flips, to store the minimum number of operations required.
  2. Traverse even indices of the array arr[] perform the following steps:
    • If the current element is odd, then continue iterating.
    • Otherwise, iterate every 2nd element starting from that index, until an even element is encountered. After complete traversal of the array or if an even element is encountered, increment flips by 1.
  3. Repeat step 2 for odd indices also.
  4. After completing the above steps, print the value of flips as the minimum number of operations as required.

Below is the implementation of the above approach:  

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count minimum subarrays
// whose odd-indexed elements need to
// be incremented to make array odd
void minOperations(int arr[], int n)
{
    // Stores the minimum number of
    // operations required
    int flips = 0;
 
    // Iterate over even-indices
    for (int i = 0; i < n; i += 2) {
 
        // Check if the current
        // element is odd
        if (arr[i] % 2 == 1) {
 
            // If true, continue
            continue;
        }
 
        // Otherwise, mark the starting
        // of the subarray and iterate
        // until i < n and arr[i] is even
        while (i < n && arr[i] % 2 == 0) {
            i += 2;
        }
 
        // Increment number of operations
        flips++;
    }
 
    // Iterate over odd indexed
    // positions of arr[]
    for (int i = 1; i < n; i += 2) {
 
        // Check if the current
        // element is odd
        if (arr[i] % 2 == 1) {
 
            // If true, continue
            continue;
        }
 
        // Otherwise, mark the starting
        // of the subarray and iterate
        // until i < n and arr[i] is even
        while (i < n && arr[i] % 2 == 0) {
            i += 2;
        }
 
        // Increment the number
        // of operations
        flips++;
    }
 
    // Print the number of operations
    cout << flips;
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 3, 4, 3, 5, 3, 2 };
    int N = sizeof(arr) / sizeof(int);
 
    // Function Call
    minOperations(arr, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
       
// Function to count minimum subarrays
// whose odd-indexed elements need to
// be incremented to make array odd
static void minOperations(int arr[], int n)
{
     
    // Stores the minimum number of
    // operations required
    int flips = 0;
     
    // Iterate over even-indices
    for(int i = 0; i < n; i += 2)
    {
         
        // Check if the current
        // element is odd
        if (arr[i] % 2 == 1)
        {
             
            // If true, continue
            continue;
        }
         
        // Otherwise, mark the starting
        // of the subarray and iterate
        // until i < n and arr[i] is even
        while (i < n && arr[i] % 2 == 0)
        {
            i += 2;
        }
         
        // Increment number of operations
        flips++;
    }
     
    // Iterate over odd indexed
    // positions of arr[]
    for(int i = 1; i < n; i += 2)
    {
         
        // Check if the current
        // element is odd
        if (arr[i] % 2 == 1)
        {
             
            // If true, continue
            continue;
        }
  
        // Otherwise, mark the starting
        // of the subarray and iterate
        // until i < n and arr[i] is even
        while (i < n && arr[i] % 2 == 0)
        {
            i += 2;
        }
         
        // Increment the number
        // of operations
        flips++;
    }
     
    // Print the number of operations
    System.out.println(flips);
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 2, 3, 4, 3, 5, 3, 2 };
    int N = arr.length;
     
    // Function Call
    minOperations(arr, N);
}
}
 
// This code is contributed by jana_sayantan

C#




// C# program for the above approach
using System;
class GFG
{
       
// Function to count minimum subarrays
// whose odd-indexed elements need to
// be incremented to make array odd
static void minOperations(int []arr, int n)
{
     
    // Stores the minimum number of
    // operations required
    int flips = 0;
     
    // Iterate over even-indices
    for(int i = 0; i < n; i += 2)
    {
         
        // Check if the current
        // element is odd
        if (arr[i] % 2 == 1)
        {
             
            // If true, continue
            continue;
        }
         
        // Otherwise, mark the starting
        // of the subarray and iterate
        // until i < n and arr[i] is even
        while (i < n && arr[i] % 2 == 0)
        {
            i += 2;
        }
         
        // Increment number of operations
        flips++;
    }
     
    // Iterate over odd indexed
    // positions of []arr
    for(int i = 1; i < n; i += 2)
    {
         
        // Check if the current
        // element is odd
        if (arr[i] % 2 == 1)
        {
             
            // If true, continue
            continue;
        }
  
        // Otherwise, mark the starting
        // of the subarray and iterate
        // until i < n and arr[i] is even
        while (i < n && arr[i] % 2 == 0)
        {
            i += 2;
        }
         
        // Increment the number
        // of operations
        flips++;
    }
     
    // Print the number of operations
    Console.WriteLine(flips);
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 2, 3, 4, 3, 5, 3, 2 };
    int N = arr.Length;
     
    // Function Call
    minOperations(arr, N);
}
}
 
// This code is contributed by Princi Singh

Python3




# Python3 program for the above approach
 
# Function to count minimum subarrays
# whose odd-indexed elements need to
# be incremented to make array odd
def minOperations(arr, n) :
 
    # Stores the minimum number of
    # operations required
    flips = 0;
    i = 0;
     
    # Iterate over even-indices
    while i < n :
 
        # Check if the current
        # element is odd
        if (arr[i] % 2 == 1) :           
            i += 2;
             
            # If true, continue
            continue;
 
        # Otherwise, mark the starting
        # of the subarray and iterate
        # until i < n and arr[i] is even
        while (i < n and arr[i] % 2 == 0) :
            i += 2;
 
        # Increment number of operations
        flips += 1;       
        i += 2;
 
    # Iterate over odd indexed
    # positions of arr[]
    i = 1
    while i < n :
 
        # Check if the current
        # element is odd
        if (arr[i] % 2 == 1) :
            i += 2;
             
            # If true, continue
            continue;
 
        # Otherwise, mark the starting
        # of the subarray and iterate
        # until i < n and arr[i] is even
        while (i < n and arr[i] % 2 == 0) :
            i += 2;
 
        # Increment the number
        # of operations
        flips += 1;       
        i += 2;
 
    # Print the number of operations
    print(flips);
 
# Driver Code
if __name__ == "__main__" :
 
    arr = [ 2, 3, 4, 3, 5, 3, 2 ];
    N = len(arr);
 
    # Function Call
    minOperations(arr, N);
 
    # This code is contributed by AnkThon

Javascript




<script>
// Javascript program for the above approach
 
// Function to count minimum subarrays
// whose odd-indexed elements need to
// be incremented to make array odd
function minOperations(arr, n)
{
      
    // Stores the minimum number of
    // operations required
    let flips = 0;
      
    // Iterate over even-indices
    for(let i = 0; i < n; i += 2)
    {
          
        // Check if the current
        // element is odd
        if (arr[i] % 2 == 1)
        {
              
            // If true, continue
            continue;
        }
          
        // Otherwise, mark the starting
        // of the subarray and iterate
        // until i < n and arr[i] is even
        while (i < n && arr[i] % 2 == 0)
        {
            i += 2;
        }
          
        // Increment number of operations
        flips++;
    }
      
    // Iterate over odd indexed
    // positions of arr[]
    for(let i = 1; i < n; i += 2)
    {
          
        // Check if the current
        // element is odd
        if (arr[i] % 2 == 1)
        {
              
            // If true, continue
            continue;
        }
   
        // Otherwise, mark the starting
        // of the subarray and iterate
        // until i < n and arr[i] is even
        while (i < n && arr[i] % 2 == 0)
        {
            i += 2;
        }
          
        // Increment the number
        // of operations
        flips++;
    }
      
    // Prlet the number of operations
    document.write(flips);
}
 
    // Driver Code
     
    let arr = [ 2, 3, 4, 3, 5, 3, 2 ];
    let N = arr.length;
      
    // Function Call
    minOperations(arr, N);
     
// This code is contributed by souravghosh0416.
</script>
Output: 
2

 

Time Complexity: O(N)
Auxiliary Space: O(1)

 

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