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Minimize the non-zero elements in the Array by given operation
  • Last Updated : 01 Jun, 2020

Given an array arr[] of length N, the task is to minimize the count of the number of non-zero elements by adding the value of the current element to any of its adjacent element and subtracting from the current element at most once.

Examples:

Input: arr[] = { 1, 0, 1, 0, 0, 1 }
Output: 2
Explanation:
Operation 1: arr[0] -> arr[1], arr[] = {0, 1, 1, 0, 0, 1}
Operation 2: arr[2] -> arr[1], arr[] = {0, 2, 0, 0, 0, 1}
Count of non-zero elements = 2

Input: arr[] = { 1, 0, 1, 1, 1, 0, 1 }
Output: 3
Explanation:
Operation 1: arr[2] -> arr[3], arr[] = {1, 0, 0, 2, 1, 0, 1}
Operation 2: arr[4] -> arr[3], arr[] = {1, 0, 0, 3, 0, 0, 1}
Count of non-zero elements = 3

Approach: The idea is to use greedy algorithms to choose greedily at each step. The key observation in the problem is there can be only three possibilities of three consecutive indices i, j and k, i.e.



  • Both the end index have non-zero element.
  • Any one of the index have non-zero element.

In the above two cases, we can always add the values to the middle element and substract to the adjacent elements. Similarly, we can greedily choose the operations and update the elements of the array to minimize the non-zero elements.

Below is the implementation of the above approach:

C++

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// C++ implementation to minimize the
// non-zero elements in the array
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to minimize the non-zero
// elements in the given array
int minOccupiedPosition(int A[], int n)
{
  
    // To store the min pos needed
    int minPos = 0;
  
    // Loop to iterate over the elements
    // of the given array
    for (int i = 0; i < n; ++i) {
  
        // If current position A[i] is occupied
        // the we can place A[i], A[i+1] and A[i+2]
        // elements together at A[i+1] if exists.
        if (A[i] > 0) {
            ++minPos;
            i += 2;
        }
    }
  
    return minPos;
}
  
// Driver Code
int main()
{
    int A[] = { 8, 0, 7, 0, 0, 6 };
    int n = sizeof(A) / sizeof(A[0]);
  
    // Function Call
    cout << minOccupiedPosition(A, n);
    return 0;
}

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Java

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// Java implementation to minimize the
// non-zero elements in the array
  
class GFG{
  
// Function to minimize the non-zero
// elements in the given array
static int minOccupiedPosition(int A[], int n)
{
      
    // To store the min pos needed
    int minPos = 0;
  
    // Loop to iterate over the elements
    // of the given array
    for (int i = 0; i < n; ++i)
    {
  
        // If current position A[i] is occupied
        // the we can place A[i], A[i+1] and A[i+2]
        // elements together at A[i+1] if exists.
        if (A[i] > 0) {
            ++minPos;
            i += 2;
        }
    }
    return minPos;
}
  
// Driver Code
public static void main(String[] args)
{
    int A[] = { 8, 0, 7, 0, 0, 6 };
    int n = A.length;
  
    // Function Call
    System.out.print(minOccupiedPosition(A, n));
}
}
  
// This code is contributed by gauravrajput1

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Python3

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# Python3 implementation to minimize the
# non-zero elements in the array
  
# Function to minimize the non-zero
# elements in the given array
def minOccupiedPosition(A, n):
  
    # To store the min pos needed
    minPos = 0
  
    # Loop to iterate over the elements
    # of the given array
    i = 0
    while i < n:
  
        # If current position A[i] is
        # occupied the we can place A[i],
        # A[i+1] and A[i+2] elements
        # together at A[i+1] if exists.
        if(A[i] > 0):
  
            minPos += 1
            i += 2
        i += 1
          
    return minPos
  
# Driver Code
if __name__ == '__main__':
      
    A = [ 8, 0, 7, 0, 0, 6 ]
    n = len(A)
  
    # Function Call 
    print(minOccupiedPosition(A, n))
      
# This code is contributed by Shivam Singh

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C#

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// C# implementation to minimize the 
// non-zero elements in the array 
using System;
  
class GFG {
      
// Function to minimize the non-zero 
// elements in the given array 
static int minOccupiedPosition(int[] A, int n) 
          
    // To store the min pos needed 
    int minPos = 0; 
      
    // Loop to iterate over the elements 
    // of the given array 
    for(int i = 0; i < n; ++i) 
    
          
       // If current position A[i] is occupied 
       // the we can place A[i], A[i+1] and A[i+2] 
       // elements together at A[i+1] if exists. 
       if (A[i] > 0) 
       
           ++minPos; 
           i += 2; 
       
    
    return minPos; 
  
// Driver code    
static void Main() 
{
    int[] A = { 8, 0, 7, 0, 0, 6 }; 
    int n = A.Length; 
  
    // Function Call 
    Console.WriteLine(minOccupiedPosition(A, n));
}
}
  
// This code is contributed by divyeshrabadiya07

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Output:

2

Time Complexity: O(N).

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