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 subtract 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++
// 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 arrayint 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 Codeint main(){ int A[] = { 8, 0, 7, 0, 0, 6 }; int n = sizeof(A) / sizeof(A[0]); // Function Call cout << minOccupiedPosition(A, n); return 0;} |
Java
// Java implementation to minimize the// non-zero elements in the arrayclass GFG{// Function to minimize the non-zero// elements in the given arraystatic 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 Codepublic 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 |
Python3
# Python3 implementation to minimize the# non-zero elements in the array# Function to minimize the non-zero# elements in the given arraydef 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 Codeif __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 |
C#
// 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 |
Javascript
<script>// Javascript implementation to minimize the// non-zero elements in the array// Function to minimize the non-zero// elements in the given arrayfunction minOccupiedPosition( A, n){ // To store the min pos needed var minPos = 0; // Loop to iterate over the elements // of the given array for (var 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;} var A = [ 8, 0, 7, 0, 0, 6 ]; var n = A.length; // Function Call document.write(minOccupiedPosition(A, n));// This code is contributed by SoumikMondal</script> |
2
Time Complexity: O(N).
Auxiliary Space: O(1)
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