Minimize the non-zero elements in the Array by given operation

• Last Updated : 10 Nov, 2021

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:
Explanation:
Operation 1: arr -> arr, arr[] = {0, 1, 1, 0, 0, 1}
Operation 2: arr -> arr, arr[] = {0, 2, 0, 0, 0, 1}
Count of non-zero elements = 2
Input: arr[] = { 1, 0, 1, 1, 1, 0, 1 }
Output:
Explanation:
Operation 1: arr -> arr, arr[] = {1, 0, 0, 2, 1, 0, 1}
Operation 2: arr -> arr, 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 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);     // Function Call    cout << minOccupiedPosition(A, n);    return 0;}

Java

 // Java implementation to minimize the// non-zero elements in the array class 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 arrayusing System; class 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 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


Output:
2

Time Complexity: O(N).

Auxiliary Space: O(1)

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