Given an array arr[] consisting of N non-negative integers, the task is to find the minimum number of subarrays that needs to be reduced by 1 such that all the array elements are equal to 0.
Example:
Input: arr[] = {1, 2, 3, 2, 1}
Output: 3
Explanation:
Operation 1: {1, 2, 3, 2, 1} -> {0, 1, 2, 1, 0}
Operation 2: {0, 1, 2, 1, 0} -> {0, 0, 1, 0, 0}
Operation 3: {0, 0, 1, 0, 0} -> {0, 0, 0, 0, 0}Input: arr[] = {5, 4, 3, 4, 4}
Output: 6
Explanation:
{5, 4, 3, 4, 4} -> {4, 3, 2, 3, 3} -> {3, 2, 1, 2, 2} -> {2, 1, 0, 1, 1} -> {2, 1, 0, 0, 0} -> {1, 0, 0, 0, 0} -> {0, 0, 0, 0, 0}
Approach:
This can be optimally done by traversing the given array from index 0, finding the answer up to index i, where 0 ? i < N. If arr[i] ? arr[i+1], then (i + 1)th element can be included in every subarray operation of ith element, thus requiring no extra operations. If arr[i] < arr[i + 1], then (i + 1)th element can be included in every subarray operation of ith element and after all operations, arr[i+1] becomes arr[i+1]-arr[i]. Therefore, we need arr[i+1]-arr[i] extra operations to reduce it zero.
Follow the below steps to solve the problem:
- Add the first element arr[0] to answer as we need at least arr[0] to make the given array 0.
- Traverse over indices [1, N-1] and for every element, check if it is greater than the previous element. If found to be true, add their difference to the answer.
Below is the implementation of above approach:
C++
// C++ Program to implement // the above approach #include <bits/stdc++.h> using namespace std; // Function to count the minimum // number of subarrays that are // required to be decremented by 1 int min_operations(vector< int >& A) { // Base Case if (A.size() == 0) return 0; // Initialize ans to first element int ans = A[0]; for ( int i = 1; i < A.size(); i++) { // For A[i] > A[i-1], operation // (A[i] - A[i - 1]) is required ans += max(A[i] - A[i - 1], 0); } // Return the answer return ans; } // Driver Code int main() { vector< int > A{ 1, 2, 3, 2, 1 }; cout << min_operations(A) << "\n" ; return 0; } |
Java
// Java Program to implement // the above approach import java.io.*; class GFG { // Function to count the minimum // number of subarrays that are // required to be decremented by 1 static int min_operations( int A[], int n) { // Base Case if (n == 0 ) return 0 ; // Initializing ans to first element int ans = A[ 0 ]; for ( int i = 1 ; i < n; i++) { // For A[i] > A[i-1], operation // (A[i] - A[i - 1]) is required if (A[i] > A[i - 1 ]) { ans += A[i] - A[i - 1 ]; } } // Return the count return ans; } // Driver Code public static void main(String[] args) { int n = 5 ; int A[] = { 1 , 2 , 3 , 2 , 1 }; System.out.println(min_operations(A, n)); } } |
Python
# Python Program to implement # the above approach # Function to count the minimum # number of subarrays that are # required to be decremented by 1 def min_operations(A): # Base case if len (A) = = 0 : return 0 # Initializing ans to first element ans = A[ 0 ] for i in range ( 1 , len (A)): if A[i] > A[i - 1 ]: ans + = A[i] - A[i - 1 ] return ans # Driver Code A = [ 1 , 2 , 3 , 2 , 1 ] print (min_operations(A)) |
C#
// C# program to implement // the above approach using System; class GFG{ // Function to count the minimum // number of subarrays that are // required to be decremented by 1 static int min_operations( int [] A, int n) { // Base Case if (n == 0) return 0; // Initializing ans to first element int ans = A[0]; for ( int i = 1; i < n; i++) { // For A[i] > A[i-1], operation // (A[i] - A[i - 1]) is required if (A[i] > A[i - 1]) { ans += A[i] - A[i - 1]; } } // Return the count return ans; } // Driver Code public static void Main() { int n = 5; int [] A = { 1, 2, 3, 2, 1 }; Console.WriteLine(min_operations(A, n)); } } // This code is contributed by bolliranadheer |
Javascript
<script> // Javascript program to implement // the above approach // Function to count the minimum // number of subarrays that are // required to be decremented by 1 function min_operations(A) { // Base Case if (A.length == 0) return 0; // Initialize ans to first element let ans = A[0]; for (let i = 1; i < A.length; i++) { // For A[i] > A[i-1], operation // (A[i] - A[i - 1]) is required ans += Math.max(A[i] - A[i - 1], 0); } // Return the answer return ans; } // Driver Code let A = [ 1, 2, 3, 2, 1 ]; document.write(min_operations(A)); // This code is contributed by subhammahato348 </script> |
Output:
3
Time Complexity: O(N)
Auxiliary Space: O(1)
Related Topic: Subarrays, Subsequences, and Subsets in Array
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