Minimum increment/decrement to make array non-Increasing

• Difficulty Level : Hard
• Last Updated : 29 Oct, 2021

Given an array a, your task is to convert it into a non-increasing form such that we can either increment or decrement the array value by 1 in minimum changes possible.

Examples :

Input : a[] = {3, 1, 2, 1}
Output : 1
Explanation :
We can convert the array into 3 1 1 1 by
changing 3rd element of array i.e. 2
into its previous integer 1 in one step
hence only one step is required.

Input : a[] = {3, 1, 5, 1}
Output : 4
We need to decrease 5 to 1 to make array sorted
in non-increasing order.

Input : a[] = {1, 5, 5, 5}
Output : 4
We need to increase 1 to 5.

Brute-Force approach: We consider both possibilities for every element and find a minimum of two possibilities.

Efficient Approach: Calculate the sum of absolute differences between the final array elements and the current array elements. Thus, the answer will be the sum of the difference between the ith element and the smallest element occurred until then. For this, we can maintain a min-heap to find the smallest element encountered till then. In the min-priority queue, we will put the elements and new elements are compared with the previous minimum. If new minimum is found we will update it, this is done because each of the next element which is coming should be smaller than the current minimum element found till now. Here, we calculate the difference so that we can get how much we have to change the current number so that it will be equal or less than previous numbers encountered till. Lastly, the sum of all these differences will be our answer as this will give the final value up to which we have to change the elements.

Below is the implementation of the above approach:

C++

 // CPP code to count the change required to// convert the array into non-increasing array#include using namespace std; int DecreasingArray(int a[], int n){    int sum = 0, dif = 0;     // min heap    priority_queue, greater > pq;     // Here in the loop we will    // check that whether the upcoming    // element of array is less than top    // of priority queue. If yes then we    // calculate the difference. After    // that we will remove that element    // and push the current element in    // queue. And the sum is incremented    // by the value of difference    for (int i = 0; i < n; i++) {        if (!pq.empty() && pq.top() < a[i]) {            dif = a[i] - pq.top();            sum += dif;            pq.pop();        }        pq.push(a[i]);    }     return sum;} // Driver Codeint main(){    int a[] = { 3, 1, 2, 1 };    int n = sizeof(a) / sizeof(a);     cout << DecreasingArray(a, n);     return 0;}

Java

 // Java code to count the change required to// convert the array into non-increasing arrayimport java.util.PriorityQueue; class GFG{    public static int DecreasingArray(int a[], int n)    {        int sum = 0, dif = 0;         PriorityQueue pq = new PriorityQueue<>();         // Here in the loop we will        // check that whether the upcoming        // element of array is less than top        // of priority queue. If yes then we        // calculate the difference. After        // that we will remove that element        // and push the current element in        // queue. And the sum is incremented        // by the value of difference        for (int i = 0; i < n; i++)        {            if (!pq.isEmpty() && pq.element() < a[i])            {                dif = a[i] - pq.element();                sum += dif;                pq.remove();            }            pq.add(a[i]);        }         return sum;    }     // Driver Code    public static void main(String[] args)    {                 int[] a = {3, 1, 2, 1};                 int n = a.length;         System.out.println(DecreasingArray(a, n));    }} // This Code is contributed by sanjeev2552

Python3

 # Python3 code to count the change required to# convert the array into non-increasing arrayfrom queue import PriorityQueue def DecreasingArray(a, n):         ss, dif = (0,0)         # min heap    pq = PriorityQueue()     # Here in the loop we will    # check that whether the upcoming    # element of array is less than top    # of priority queue. If yes then we    # calculate the difference. After    # that we will remove that element    # and push the current element in    # queue. And the sum is incremented    # by the value of difference    for i in range(n):        tmp = 0                 if not pq.empty():            tmp = pq.get()            pq.put(tmp)                 if not pq.empty() and tmp < a[i]:            dif = a[i] - tmp            ss += dif            pq.get()                 pq.put(a[i])           return ss     # Driver code   if __name__=="__main__":         a = [ 3, 1, 2, 1 ]    n = len(a)      print(DecreasingArray(a, n))     # This code is contributed by rutvik_56

C#

 // C# code to count the change required to// convert the array into non-increasing arrayusing System;using System.Collections.Generic;class GFG{    static int DecreasingArray(int[] a, int n)    {        int sum = 0, dif = 0;              // min heap        List pq = new List();              // Here in the loop we will        // check that whether the upcoming        // element of array is less than top        // of priority queue. If yes then we        // calculate the difference. After        // that we will remove that element        // and push the current element in        // queue. And the sum is incremented        // by the value of difference        for (int i = 0; i < n; i++)        {            if (pq.Count > 0 && pq < a[i])            {                dif = a[i] - pq;                sum += dif;                pq.RemoveAt(0);            }            pq.Add(a[i]);            pq.Sort();        }              return sum;    }     // Driver code  static void Main()  {    int[] a = { 3, 1, 2, 1 };    int n = a.Length;      Console.Write(DecreasingArray(a, n));  }} // This code is contributed by divyeshrabadiya07.

Javascript


Output:
1

Time Complexity: O(n log(n))
Space Complexity: O(n)
Also see : Convert to strictly increasing array with minimum changes.

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