Given an array of positive distinct integers. We need to find the only element whose replacement with any other value makes array elements distinct consecutive. If it is not possible to make array elements consecutive, return -1.
Examples :
Input : arr[] = {45, 42, 46, 48, 47}
Output : 42
Explanation: We can replace 42 with either
44 or 48 to make array consecutive.
Input : arr[] = {5, 6, 7, 9, 10}
Output : 5 [OR 10]
Explanation: We can either replace 5 with 8
or 10 with 8 to make array elements
consecutive.
Input : arr[] = {5, 6, 7, 9, 8}
Output : Array elements are already consecutive
A Naive Approach is to check each element of arr[], after replacing of which makes consecutive or not. Time complexity for this approach O(n2)
A Better Approach is based on an important observation that either the smallest or the largest element would be answer if answer exists. If answer exists, then there are two cases.
1) Series of consecutive elements starts with minimum element of array then continues by adding 1 to previous.
2) Series of consecutive elements start with maximum element of array, then continues by subtracting 1 from previous.
We make above two series and for every series, we search series elements in array. If for both series, number of mismatches are more than 1, then answer does not exist. If any series is found with one mismatch, then we have answer.
C++
// C++ program to find an element replacement// of which makes the array elements consecutive.#include <bits/stdc++.h>using namespace std;int findElement(int arr[], int n){ sort(arr, arr+n); // Making a series starting from first element // and adding 1 to every element. int mismatch_count1 = 0, res; int next_element = arr[n-1] - n + 1; for (int i=0; i<n-1; i++) { if (binary_search(arr, arr+n, next_element) == 0) { res = arr[0]; mismatch_count1++; } next_element++; } // If only one mismatch is found. if (mismatch_count1 == 1) return res; // If no mismatch found, elements are // already consecutive. if (mismatch_count1 == 0) return 0; // Making a series starting from last element // and subtracting 1 to every element. int mismatch_count2 = 0; next_element = arr[0] + n - 1; for (int i=n-1; i>=1; i--) { if (binary_search(arr, arr+n, next_element) == 0) { res = arr[n-1]; mismatch_count2++; } next_element--; } // If only one mismatch is found. if (mismatch_count2 == 1) return res; return -1; }// Driver codeint main(){ int arr[] = {7, 5, 12, 8} ; int n = sizeof(arr)/sizeof(arr[0]); int res = findElement(arr,n); if (res == -1) cout << "Answer does not exist"; else if (res == 0) cout << "Elements are already consecutive"; else cout << res; return 0;} |
Java
// Java program to find an element // replacement of which makes // the array elements consecutive.import java.io.*;import java.util.Arrays;class GFG{ static int findElement(int []arr, int n) { Arrays.sort(arr); // Making a series starting // from first element and // adding 1 to every element. int mismatch_count1 = 0, res = 0; int next_element = arr[n - 1] - n + 1; for (int i = 0; i < n - 1; i++) { if (Arrays.binarySearch(arr, next_element) < 0) { res = arr[0]; mismatch_count1++; } next_element++; } // If only one mismatch is found. if (mismatch_count1 == 1) return res; // If no mismatch found, elements // are already consecutive. if (mismatch_count1 == 0) return 0; // Making a series starting // from last element and // subtracting 1 to every element. int mismatch_count2 = 0; next_element = arr[0] + n - 1; for (int i = n - 1; i >= 1; i--) { if (Arrays.binarySearch(arr, next_element) < 0) { res = arr[n - 1]; mismatch_count2++; } next_element--; } // If only one mismatch is found. if (mismatch_count2 == 1) return res; return -1; } // Driver code public static void main(String args[]) { int []arr = new int[]{7, 5, 12, 8} ; int n = arr.length; int res = findElement(arr,n); if (res == -1) System.out.print("Answer does not exist"); else if (res == 0) System.out.print("Elements are " + "already consecutive"); else System.out.print(res); }}// This code is contributed by // Manish Shaw(manishshaw1) |
Python3
# Python3 program to find an element# replacement of which makes the # array elements consecutive. from bisect import bisect_left def BinarySearch(a, x): i = bisect_left(a, x) if i != len(a) and a[i] == x: return i else: return -1 def findElement(arr, n): arr.sort() # Making a series starting # from first element and # adding 1 to every element. mismatch_count1 = 0 res = 0 next_element = arr[n - 1] - n + 1 for i in range(n - 1): if (BinarySearch(arr, next_element) == -1): res = arr[0] mismatch_count1 += 1 next_element += 1 # If only one mismatch is found. if (mismatch_count1 == 1): return res # If no mismatch found, elements are # already consecutive. if (mismatch_count1 == 0): return 0 # Making a series starting from last element # and subtracting 1 to every element. mismatch_count2 = 0 next_element = arr[0] + n - 1 for i in range(n - 1, 0, -1): if BinarySearch(arr, next_element) == -1: res = arr[n - 1] mismatch_count2 += 1 next_element -= 1 # If only one mismatch is found. if(mismatch_count2 == 1): return res return -1# Driver codeif __name__=="__main__": arr = [ 7, 5, 12, 8 ] n = len(arr) res = findElement(arr, n) if (res == -1): print("Answer does not exist") elif (res == 0): print("Elements are already consecutive") else: print(res) # This code is contributed by rutvik_56 |
C#
// C# program to find an element // replacement of which makes // the array elements consecutive.using System;using System.Linq;using System.Collections.Generic;class GFG{ static int findElement(int []arr, int n) { Array.Sort(arr); // Making a series starting // from first element and // adding 1 to every element. int mismatch_count1 = 0, res = 0; int next_element = arr[n - 1] - n + 1; for (int i = 0; i < n - 1; i++) { if (Array.BinarySearch(arr, next_element) < 0) { res = arr[0]; mismatch_count1++; } next_element++; } // If only one mismatch is found. if (mismatch_count1 == 1) return res; // If no mismatch found, elements // are already consecutive. if (mismatch_count1 == 0) return 0; // Making a series starting // from last element and // subtracting 1 to every element. int mismatch_count2 = 0; next_element = arr[0] + n - 1; for (int i = n - 1; i >= 1; i--) { if (Array.BinarySearch(arr, next_element) < 0) { res = arr[n - 1]; mismatch_count2++; } next_element--; } // If only one mismatch is found. if (mismatch_count2 == 1) return res; return -1; } // Driver code static void Main() { int []arr = new int[]{7, 5, 12, 8} ; int n = arr.Length; int res = findElement(arr,n); if (res == -1) Console.Write("Answer does not exist"); else if (res == 0) Console.Write("Elements are " + "already consecutive"); else Console.Write(res); }}// This code is contributed by // Manish Shaw(manishshaw1) |
Javascript
<script>// JavaScript program to find an element replacement// of which makes the array elements consecutive.function binary_search(arr, x) { let start=0, end=arr.length-1; // Iterate while start not meets end while (start<=end){ // Find the mid index let mid=Math.floor((start + end)/2); // If element is present at mid, return True if (arr[mid]===x) return mid; // Else look in left or right half accordingly else if (arr[mid] < x) start = mid + 1; else end = mid - 1; } return -1;}function findElement( arr, n){ arr.sort(function(a,b){return a-b}); // Making a series starting from first element // and adding 1 to every element. let mismatch_count1 = 0, res; let next_element = arr[n-1] - n + 1; for (let i=0; i<n-1; i++) { if (binary_search(arr, next_element) < 0) { res = arr[0]; mismatch_count1++; } next_element++; } // If only one mismatch is found. if (mismatch_count1 == 1) return res; // If no mismatch found, elements are // already consecutive. if (mismatch_count1 == 0) return 0; // Making a series starting from last element // and subtracting 1 to every element. let mismatch_count2 = 0; next_element = arr[0] + n - 1; for (let i=n-1; i>=1; i--) { if (binary_search(arr, next_element) < 0) { res = arr[n-1]; mismatch_count2++; } next_element--; } // If only one mismatch is found. if (mismatch_count2 == 1) return res; return -1; }// Driver code let a = [7, 5, 12, 8] ; let N = a.length; let res = findElement(a,N); if (res == -1) document.write( "Answer does not exist"); else if (res == 0) document.write( "Elements are already consecutive"); else document.write(res);</script> |
Output :
12
Time Complexity: O(n Log n)
Auxiliary Space: O(1)
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