Given an array arr[] of positive integers, the task is to find whether it is possible to make this array strictly decreasing by modifying at most one element.
Examples:
Input: arr[] = {12, 9, 10, 5, 2}
Output: Yes
{12, 11, 10, 5, 2} is one of the valid solutions.
Input: arr[] = {1, 2, 3, 4}
Output: No
Approach: For every element arr[i], if it is greater than both arr[i – 1] and arr[i + 1] or it is smaller than both arr[i – 1] and arr[i + 1] then arr[i] needs to be modified.
i.e arr[i] = (arr[i – 1] + arr[i + 1]) / 2. If after modification, arr[i] = arr[i – 1] or arr[i + 1] then the array cannot be made strictly decreasing without affecting at most one element else count all such modifications, if the count of modifications in the end is less than or equal to 1 then print Yes else print No.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function that returns true if the array// can be made strictly decreasing// with at most one changebool check(vector<int> arr, int n){ // To store the number of modifications // required to make the array // strictly decreasing int modify = 0; // Check whether the last element needs // to be modify or not if (arr[n - 1] >= arr[n - 2]) { arr[n - 1] = arr[n - 2] - 1; modify++; } // Check whether the first element needs // to be modify or not if (arr[0] <= arr[1]) { arr[0] = arr[1] + 1; modify++; } // Loop from 2nd element to the 2nd last element for (int i = n - 2; i > 0; i--) { // Check whether arr[i] needs to be modified if ((arr[i - 1] <= arr[i] && arr[i + 1] <= arr[i]) || (arr[i - 1] >= arr[i] && arr[i + 1] >= arr[i])) { // Modifying arr[i] arr[i] = (arr[i - 1] + arr[i + 1]) / 2; modify++; // Check if arr[i] is equal to any of // arr[i-1] or arr[i+1] if (arr[i] == arr[i - 1] || arr[i] == arr[i + 1]) return false; } } // If more than 1 modification is required if (modify > 1) return false; return true;}// Driver codeint main(){ vector<int> arr = { 10, 5, 11, 2 }; int n = arr.size(); if (check(arr, n)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java implementation of the approachclass GFG { // Function that returns true if the array // can be made strictly decreasing // with at most one change public static boolean check(int[] arr, int n) { // To store the number of modifications // required to make the array // strictly decreasing int modify = 0; // Check whether the last element needs // to be modify or not if (arr[n - 1] >= arr[n - 2]) { arr[n - 1] = arr[n - 2] - 1; modify++; } // Check whether the first element needs // to be modify or not if (arr[0] <= arr[1]) { arr[0] = arr[1] + 1; modify++; } // Loop from 2nd element to the 2nd last element for (int i = n - 2; i > 0; i--) { // Check whether arr[i] needs to be modified if ((arr[i - 1] <= arr[i] && arr[i + 1] <= arr[i]) || (arr[i - 1] >= arr[i] && arr[i + 1] >= arr[i])) { // Modifying arr[i] arr[i] = (arr[i - 1] + arr[i + 1]) / 2; modify++; // Check if arr[i] is equal to any of // arr[i-1] or arr[i+1] if (arr[i] == arr[i - 1] || arr[i] == arr[i + 1]) return false; } } // If more than 1 modification is required if (modify > 1) return false; return true; } // Driver code public static void main(String[] args) { int[] arr = { 10, 5, 11, 3 }; int n = arr.length; if (check(arr, n)) System.out.print("Yes"); else System.out.print("No"); }} |
Python3
# Python3 implementation of the approach# Function that returns true if the array # can be made strictly decreasing # with at most one changedef check(arr, n): modify = 0 # Check whether the last element needs # to be modify or not if (arr[n - 1] >= arr[n - 2]): arr[n-1] = arr[n-2] - 1 modify += 1 # Check whether the first element needs # to be modify or not if (arr[0] <= arr[1]): arr[0] = arr[1] + 1 modify += 1 # Loop from 2nd element to the 2nd last element for i in range(n-2, 0, -1): # Check whether arr[i] needs to be modified if (arr[i - 1] <= arr[i] and arr[i + 1] <= arr[i]) or \ (arr[i - 1] >= arr[i] and arr[i + 1] >= arr[i]): # Modifying arr[i] arr[i] = (arr[i - 1] + arr[i + 1]) // 2 modify += 1 # Check if arr[i] is equal to any of # arr[i-1] or arr[i + 1] if (arr[i] == arr[i - 1] or arr[i] == arr[i + 1]): return False # If more than 1 modification is required if (modify > 1): return False return True# Driver codeif __name__ == "__main__": arr = [10, 5, 11, 3] n = len(arr) if (check(arr, n)): print("Yes") else: print("No") |
C#
// C# implementation of the approachusing System;class GFG{ // Function that returns true if the array // can be made strictly decreasing // with at most one change public static bool check(int[]arr, int n) { // To store the number of modifications // required to make the array // strictly decreasing int modify = 0; // Check whether the last element needs // to be modify or not if (arr[n - 1] >= arr[n - 2]) { arr[n - 1] = arr[n - 2] - 1; modify++; } // Check whether the first element needs // to be modify or not if (arr[0] <= arr[1]) { arr[0] = arr[1] + 1; modify++; } // Loop from 2nd element to the 2nd last element for (int i = n - 2; i > 0; i--) { // Check whether arr[i] needs to be modified if ((arr[i - 1] <= arr[i] && arr[i + 1] <= arr[i]) || (arr[i - 1] >= arr[i] && arr[i + 1] >= arr[i])) { // Modifying arr[i] arr[i] = (arr[i - 1] + arr[i + 1]) / 2; modify++; // Check if arr[i] is equal to any of // arr[i-1] or arr[i+1] if (arr[i] == arr[i - 1] || arr[i] == arr[i + 1]) return false; } } // If more than 1 modification is required if (modify > 1) return false; return true; } // Driver code static public void Main () { int[]arr = { 10, 5, 11, 3 }; int n = arr.Length; if (check(arr, n)) Console.Write("Yes"); else Console.Write("No"); }}// This code is contributed by ajit. |
Javascript
<script> // Javascript implementation of the approach // Function that returns true if the array // can be made strictly decreasing // with at most one change function check(arr, n) { // To store the number of modifications // required to make the array // strictly decreasing let modify = 0; // Check whether the last element needs // to be modify or not if (arr[n - 1] >= arr[n - 2]) { arr[n - 1] = arr[n - 2] - 1; modify++; } // Check whether the first element needs // to be modify or not if (arr[0] <= arr[1]) { arr[0] = arr[1] + 1; modify++; } // Loop from 2nd element to the 2nd last element for (let i = n - 2; i > 0; i--) { // Check whether arr[i] needs to be modified if ((arr[i - 1] <= arr[i] && arr[i + 1] <= arr[i]) || (arr[i - 1] >= arr[i] && arr[i + 1] >= arr[i])) { // Modifying arr[i] arr[i] = parseInt((arr[i - 1] + arr[i + 1]) / 2, 10); modify++; // Check if arr[i] is equal to any of // arr[i-1] or arr[i+1] if (arr[i] == arr[i - 1] || arr[i] == arr[i + 1]) return false; } } // If more than 1 modification is required if (modify > 1) return false; return true; } let arr = [ 10, 5, 11, 3 ]; let n = arr.length; if (check(arr, n)) document.write("Yes"); else document.write("No");</script> |
Output:
Yes
Time Complexity: O(N)
Auxiliary Space: O(1)
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