Given an array arr[], the task is to find an element that can be added to the array in order to convert it to Arithmetic Progression. If it’s impossible to convert the given array into an AP, then print -1.
Examples:
Input: arr[] = {3, 7}
Output: 11
3, 7 and 11 is a finite AP sequence.Input: a[] = {4, 6, 8, 15}
Output: -1
Approach:
- Sort the array and start traversing the array element by element and note the difference between the two consecutive elements.
- If the difference for all the elements is the same then print last element + common difference.
- If the difference is different for at most one pair (arr[i – 1], arr[i]) and diff = 2 * common difference for all other elements, then print arr[i] – common difference.
- Else print -1.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include<bits/stdc++.h>using namespace std;// Function to return the number to be// addedint getNumToAdd(int arr[], int n){ sort(arr,arr+n); int d = arr[1] - arr[0]; int numToAdd = -1; bool numAdded = false; for (int i = 2; i < n; i++) { int diff = arr[i] - arr[i - 1]; // If difference of the current // consecutive elements is // different from the common // difference if (diff != d) { // If number has already been // chosen then it's not possible // to add another number if (numAdded) return -1; // If the current different is // twice the common difference // then a number can be added midway // from current and previous element if (diff == 2 * d) { numToAdd = arr[i] - d; // Number has been chosen numAdded = true; } // It's not possible to maintain // the common difference else return -1; } } // Return last element + common difference // if no element is chosen and the array // is already in AP if (numToAdd == -1) return (arr[n - 1] + d); // Else return the chosen number return numToAdd;}// Driver codeint main(){ int arr[] = { 1, 3, 5, 7, 11, 13, 15 }; int n = sizeof(arr)/sizeof(arr[0]); cout << getNumToAdd(arr, n);}// This code is contributed// by ihritik |
Java
// Java implementation of the approachimport java.util.*;public class GFG { // Function to return the number to be // added static int getNumToAdd(int arr[], int n) { Arrays.sort(arr); int d = arr[1] - arr[0]; int numToAdd = -1; boolean numAdded = false; for (int i = 2; i < n; i++) { int diff = arr[i] - arr[i - 1]; // If difference of the current // consecutive elements is // different from the common // difference if (diff != d) { // If number has already been // chosen then it's not possible // to add another number if (numAdded) return -1; // If the current different is // twice the common difference // then a number can be added midway // from current and previous element if (diff == 2 * d) { numToAdd = arr[i] - d; // Number has been chosen numAdded = true; } // It's not possible to maintain // the common difference else return -1; } } // Return last element + common difference // if no element is chosen and the array // is already in AP if (numToAdd == -1) return (arr[n - 1] + d); // Else return the chosen number return numToAdd; } // Driver code public static void main(String[] args) { int arr[] = { 1, 3, 5, 7, 11, 13, 15 }; int n = arr.length; System.out.println(getNumToAdd(arr, n)); }} |
Python3
# Python 3 implementation of the approach# Function to return the number # to be addeddef getNumToAdd(arr, n): arr.sort(reverse = False) d = arr[1] - arr[0] numToAdd = -1 numAdded = False for i in range(2, n, 1): diff = arr[i] - arr[i - 1] # If difference of the current consecutive # elements is different from the common # difference if (diff != d): # If number has already been chosen # then it's not possible to add # another number if (numAdded): return -1 # If the current different is twice # the common difference then a # number can be added midway from # current and previous element if (diff == 2 * d): numToAdd = arr[i] - d # Number has been chosen numAdded = True # It's not possible to maintain # the common difference else: return -1 # Return last element + common difference # if no element is chosen and the array # is already in AP if (numToAdd == -1): return (arr[n - 1] + d) # Else return the chosen number return numToAdd# Driver codeif __name__ == '__main__': arr = [1, 3, 5, 7, 11, 13, 15] n = len(arr) print(getNumToAdd(arr, n))# This code is contributed# mohit kumar 29 |
C#
// C# implementation of the approachusing System;public class GFG { // Function to return the number to be // added static int getNumToAdd(int []arr, int n) { Array.Sort(arr); int d = arr[1] - arr[0]; int numToAdd = -1; bool numAdded = false; for (int i = 2; i < n; i++) { int diff = arr[i] - arr[i - 1]; // If difference of the current // consecutive elements is // different from the common // difference if (diff != d) { // If number has already been // chosen then it's not possible // to add another number if (numAdded) return -1; // If the current different is // twice the common difference // then a number can be added midway // from current and previous element if (diff == 2 * d) { numToAdd = arr[i] - d; // Number has been chosen numAdded = true; } // It's not possible to maintain // the common difference else return -1; } } // Return last element + common difference // if no element is chosen and the array // is already in AP if (numToAdd == -1) return (arr[n - 1] + d); // Else return the chosen number return numToAdd; } // Driver code public static void Main() { int []arr = { 1, 3, 5, 7, 11, 13, 15 }; int n = arr.Length; Console.WriteLine(getNumToAdd(arr, n)); }}// This code is contributed// by ihritik |
PHP
<?php// PHP implementation of the approach// Function to return the number // to be addedfunction getNumToAdd($arr, $n){ sort($arr); $d = $arr[1] - $arr[0]; $numToAdd = -1; $numAdded = false; for ($i = 2; $i < $n; $i++) { $diff = $arr[$i] - $arr[$i - 1]; // If difference of the current // consecutive elements is // different from the common // difference if ($diff != $d) { // If number has already been // chosen then it's not possible // to add another number if ($numAdded) return -1; // If the current different is // twice the common difference // then a number can be added midway // from current and previous element if ($diff == 2 * $d) { $numToAdd = $arr[$i] - $d; // Number has been chosen $numAdded = true; } // It's not possible to maintain // the common difference else return -1; } } // Return last element + common difference // if no element is chosen and the array // is already in AP if ($numToAdd == -1) return ($arr[$n - 1] + $d); // Else return the chosen number return $numToAdd;}// Driver code$arr = array( 1, 3, 5, 7, 11, 13, 15 );$n = sizeof($arr);echo getNumToAdd($arr, $n);// This code is contributed by Sachin..?> |
Javascript
<script>// Javascript implementation of the approach// Function to return the number to be// addedfunction getNumToAdd(arr, n){ arr.sort(function(a, b){return a - b}); var d = arr[1] - arr[0]; var numToAdd = -1; var numAdded = false; for(var i = 2; i < n; i++) { var diff = arr[i] - arr[i - 1]; // If difference of the current // consecutive elements is // different from the common // difference if (diff != d) { // If number has already been // chosen then it's not possible // to add another number if (numAdded) return -1; // If the current different is // twice the common difference // then a number can be added midway // from current and previous element if (diff == 2 * d) { numToAdd = arr[i] - d; // Number has been chosen numAdded = true; } // It's not possible to maintain // the common difference else return -1; } } // Return last element + common difference // if no element is chosen and the array // is already in AP if (numToAdd == -1) return (arr[n - 1] + d); // Else return the chosen number return numToAdd;}// Driver codevar arr = [ 1, 3, 5, 7, 11, 13, 15 ];var n = arr.length;document.write(getNumToAdd(arr, n));// This code is contributed by Ankita saini</script> |
Output
9
Complexity Analysis:
- Time Complexity : O(n Log n)
- Space Complexity: O(1) since only using constant variable
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