Given an array arr[] consisting of N integers, the task is to find the minimum number of increment/decrement by 1 required to be performed on array elements to make all the elements of the given array arr[] in AP. If it is not possible to make the array in AP, then print “-1”.
Examples:
Input: arr[] = {19, 16, 9, 5, 0}
Output: 3
Explanation:
Following are the order of increment/decrements of array elements is made:
- Increment array element arr[0](= 19) by 1.
- Decrement array element arr[1](= 16) by 1.
- Increment array element arr[2](= 9) by 1.
After the above operations, the array arr[] modifies to {20, 15, 10, 5, 0}, which is in AP with first term 20 and common difference -5. Therefore, the total count of element is 3.
Input: arr[] = {1, 2, 3, 4, 10}
Output: -1
Approach: The given problem can be solved by finding the first term and the common difference from the first two elements and then check if all elements can be changed to the AP sequence with the given first term and common difference by simply iterating over the array. Follow the steps below to solve the problem:
- If N is less than equal to 2, then print 0, because each such sequence is an Arithmetic Progression.
- Initialize a variable say, res as N + 1 to store the answer.
- Iterate in the range [-1, 1] using the variable a and perform the following steps:
- Iterate in the range [-1, 1] using the variable b and perform the following steps:
- Initialize a variable say changes as 0 to store the count of changed elements of the array arr[].
- If a is not equal to 0, then increase the value of changes by 1.
- If b is not equal to 0, then increase the value of changes by 1.
- Initialize a variable say, orig as arr[0] + a to store the first element and diff as (arr[1] + b) – (arr[0] + a) to store the common difference of the arithmetic progression.
- Initialize a variable say, good as true to store whether the arithmetic progression sequence with first term orig and common difference diff is possible or not.
- Iterate in the range [2, N-1] using the variable i and perform the following steps:
- Initialize a variable actual as orig+i*diff to store the actual element of the arithmetic progression at index i.
- If abs(actual – arr[i]) is greater than 1, then such arithmetic progression is unreachable. Then Set the value of good to false and break the loop.
- Else, if actual is not equal to arr[i], increase the value of changes by 1.
- After iterating through the inner for loop, update the value of res to min(changes, res).
- Iterate in the range [-1, 1] using the variable b and perform the following steps:
- After completing the above steps, if res is greater than N then assign -1 to res. Otherwise, print the value of res as the answer.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the minimum number// of elements to be changed to convert// the array into an APint findMinimumMoves(int N, int arr[]){ // If N is less than 3 if (N <= 2) { return 0; } // Stores the answer int res = N + 1; // Iterate in the range [-1, 1] for (int a = -1; a <= 1; a++) { // Iterate in the range [-1, 1] for (int b = -1; b <= 1; b++) { // Stores the total changes int changes = 0; if (a != 0) { changes++; } if (b != 0) { changes++; } // Stores the first element // of the AP int orig = (arr[0] + a); // Stores the common difference // of the AP int diff = (arr[1] + b) - (arr[0] + a); // Stores whether it is // possible to convert the // array into AP bool good = true; // Iterate in the range [2, N-1] for (int i = 2; i < N; i++) { // Stores the ith element // of the AP int actual = orig + i * diff; // If abs(actual-arr[i]) // is greater than 1 if (abs(actual - arr[i]) > 1) { // Mark as false good = false; break; } // If actual is not // equal to arr[i] if (actual != arr[i]) changes++; } if (!good) continue; // Update the value of res res = min(res, changes); } } // If res is greater than N if (res > N) res = -1; // Return the value of res return res;}// Driver Codeint main(){ int arr[] = { 19, 16, 9, 5, 0 }; int N = sizeof(arr) / sizeof(arr[0]); cout << findMinimumMoves(N, arr); return 0;} |
Java
// Java program for the above approachimport java.io.*;class GFG{ // Function to find the minimum number// of elements to be changed to convert// the array into an APstatic int findMinimumMoves(int N, int arr[]){ // If N is less than 3 if (N <= 2) { return 0; } // Stores the answer int res = N + 1; // Iterate in the range [-1, 1] for(int a = -1; a <= 1; a++) { // Iterate in the range [-1, 1] for(int b = -1; b <= 1; b++) { // Stores the total changes int changes = 0; if (a != 0) { changes++; } if (b != 0) { changes++; } // Stores the first element // of the AP int orig = (arr[0] + a); // Stores the common difference // of the AP int diff = (arr[1] + b) - (arr[0] + a); // Stores whether it is // possible to convert the // array into AP boolean good = true; // Iterate in the range [2, N-1] for(int i = 2; i < N; i++) { // Stores the ith element // of the AP int actual = orig + i * diff; // If abs(actual-arr[i]) // is greater than 1 if (Math.abs(actual - arr[i]) > 1) { // Mark as false good = false; break; } // If actual is not // equal to arr[i] if (actual != arr[i]) changes++; } if (!good) continue; // Update the value of res res = Math.min(res, changes); } } // If res is greater than N if (res > N) res = -1; // Return the value of res return res;}// Driver Codepublic static void main(String[] args){ int arr[] = { 19, 16, 9, 5, 0 }; int N = arr.length; System.out.println(findMinimumMoves(N, arr));}}// This code is contributed by Potta Lokesh |
Python3
# Python3 program for the above approach# Function to find the minimum number# of elements to be changed to convert# the array into an APdef findMinimumMoves(N, arr): # If N is less than 3 if (N <= 2): return 0 # Stores the answer res = N + 1 # Iterate in the range [-1, 1] for a in range(-1, 2, 1): # Iterate in the range [-1, 1] for b in range(-1, 2, 1): # Stores the total changes changes = 0 if (a != 0): changes += 1 if (b != 0): changes += 1 # Stores the first element # of the AP orig = (arr[0] + a) # Stores the common difference # of the AP diff = (arr[1] + b) - (arr[0] + a) # Stores whether it is # possible to convert the # array into AP good = True # Iterate in the range [2, N-1] for i in range(2, N, 1): # Stores the ith element # of the AP actual = orig + i * diff # If abs(actual-arr[i]) # is greater than 1 if (abs(actual - arr[i]) > 1): # Mark as false good = False break # If actual is not # equal to arr[i] if (actual != arr[i]): changes += 1 if (good == False): continue # Update the value of res res = min(res, changes) # If res is greater than N if (res > N): res = -1 # Return the value of res return res# Driver Codeif __name__ == '__main__': arr = [ 19, 16, 9, 5, 0 ] N = len(arr) print(findMinimumMoves(N, arr))# This code is contributed by ipg2016107 |
C#
// C# program for the above approachusing System;class GFG{ // Function to find the minimum number// of elements to be changed to convert// the array into an APstatic int findMinimumMoves(int N, int[] arr){ // If N is less than 3 if (N <= 2) { return 0; } // Stores the answer int res = N + 1; // Iterate in the range [-1, 1] for(int a = -1; a <= 1; a++) { // Iterate in the range [-1, 1] for(int b = -1; b <= 1; b++) { // Stores the total changes int changes = 0; if (a != 0) { changes++; } if (b != 0) { changes++; } // Stores the first element // of the AP int orig = (arr[0] + a); // Stores the common difference // of the AP int diff = (arr[1] + b) - (arr[0] + a); // Stores whether it is // possible to convert the // array into AP bool good = true; // Iterate in the range [2, N-1] for(int i = 2; i < N; i++) { // Stores the ith element // of the AP int actual = orig + i * diff; // If abs(actual-arr[i]) // is greater than 1 if (Math.Abs(actual - arr[i]) > 1) { // Mark as false good = false; break; } // If actual is not // equal to arr[i] if (actual != arr[i]) changes++; } if (!good) continue; // Update the value of res res = Math.Min(res, changes); } } // If res is greater than N if (res > N) res = -1; // Return the value of res return res;}// Driver codestatic public void Main(){ int[] arr = { 19, 16, 9, 5, 0 }; int N = arr.Length; Console.WriteLine(findMinimumMoves(N, arr));}}// This code is contributed by target_2. |
Javascript
<script>// Javascript program for the above approach// Function to find the minimum number// of elements to be changed to convert// the array into an APfunction findMinimumMoves(N, arr){ // If N is less than 3 if (N <= 2) { return 0; } // Stores the answer let res = N + 1; // Iterate in the range [-1, 1] for(let a = -1; a <= 1; a++) { // Iterate in the range [-1, 1] for(let b = -1; b <= 1; b++) { // Stores the total changes let changes = 0; if (a != 0) { changes++; } if (b != 0) { changes++; } // Stores the first element // of the AP let orig = (arr[0] + a); // Stores the common difference // of the AP let diff = (arr[1] + b) - (arr[0] + a); // Stores whether it is // possible to convert the // array into AP let good = true; // Iterate in the range [2, N-1] for(let i = 2; i < N; i++) { // Stores the ith element // of the AP let actual = orig + i * diff; // If abs(actual-arr[i]) // is greater than 1 if (Math.abs(actual - arr[i]) > 1) { // Mark as false good = false; break; } // If actual is not // equal to arr[i] if (actual != arr[i]) changes++; } if (!good) continue; // Update the value of res res = Math.min(res, changes); } } // If res is greater than N if (res > N) res = -1; // Return the value of res return res;}// Driver Codelet arr = [ 19, 16, 9, 5, 0 ];let N = arr.length;document.write(findMinimumMoves(N, arr));// This code is contributed by target_2 </script> |
Output:
3
Time Complexity: O(N)
Auxiliary Space: O(1)
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