k-th missing element in sorted array
Given an increasing sequence a[], we need to find the K-th missing contiguous element in the increasing sequence which is not present in the sequence. If no k-th missing element is there output -1.
Examples :
Input : a[] = {2, 3, 5, 9, 10};
k = 1;
Output : 4
Explanation: Missing Element in the increasing
sequence are {4, 6, 7, 8}. So k-th missing element
is 4
Input : a[] = {2, 3, 5, 9, 10, 11, 12};
k = 4;
Output : 8
Explanation: missing element in the increasing
sequence are {4, 6, 7, 8} so k-th missing
element is 8
Approach: Start iterating over the array elements, and for every element check if next element is consecutive or not, if not, then take difference between these two, and check if difference is greater than or equal to given k, then calculate ans = a[i] + count, else iterate for next element.
C++
#include <bits/stdc++.h> using namespace std; // Function to find k-th // missing element int missingK(int a[], int k, int n) { int difference = 0, ans = 0, count = k; bool flag = 0; // interating over the array for(int i = 0 ; i < n - 1; i++) { difference = 0; // check if i-th and // (i + 1)-th element // are not consecutive if ((a[i] + 1) != a[i + 1]) { // save their difference difference += (a[i + 1] - a[i]) - 1; // check for difference // and given k if (difference >= count) { ans = a[i] + count; flag = 1; break; } else count -= difference; } } // if found if(flag) return ans; else return -1; } // Driver code int main() { // Input array int a[] = {1, 5, 11, 19}; // k-th missing element // to be found in the array int k = 11; int n = sizeof(a) / sizeof(a[0]); // calling function to // find missing element int missing = missingK(a, k, n); cout << missing << endl; return 0; } |
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Java
// Java program to check for // even or odd import java.io.*; import java.util.*; public class GFG { // Function to find k-th // missing element static int missingK(int []a, int k, int n) { int difference = 0, ans = 0, count = k; boolean flag = false; // interating over the array for(int i = 0 ; i < n - 1; i++) { difference = 0; // check if i-th and // (i + 1)-th element // are not consecutive if ((a[i] + 1) != a[i + 1]) { // save their difference difference += (a[i + 1] - a[i]) - 1; // check for difference // and given k if (difference >= count) { ans = a[i] + count; flag = true; break; } else count -= difference; } } // if found if(flag) return ans; else return -1; } // Driver code public static void main(String args[]) { // Input array int []a = {1, 5, 11, 19}; // k-th missing element // to be found in the array int k = 11; int n = a.length; // calling function to // find missing element int missing = missingK(a, k, n); System.out.print(missing); } } // This code is contributed by // Manish Shaw (manishshaw1) |
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Python3
# Function to find k-th # missing element def missingK(a, k, n) : difference = 0 ans = 0 count = k flag = 0 # interating over the array for i in range (0, n-1) : difference = 0 # check if i-th and # (i + 1)-th element # are not consecutive if ((a[i] + 1) != a[i + 1]) : # save their difference difference += (a[i + 1] - a[i]) - 1 # check for difference # and given k if (difference >= count) : ans = a[i] + count flag = 1 break else : count -= difference # if found if(flag) : return ans else : return -1 # Driver code # Input array a = [1, 5, 11, 19] # k-th missing element # to be found in the array k = 11n = len(a) # calling function to # find missing element missing = missingK(a, k, n) print(missing) # This code is contributed by # Manish Shaw (manishshaw1) |
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C#
// C# program to check for // even or odd using System; using System.Collections.Generic; class GFG { // Function to find k-th // missing element static int missingK(int []a, int k, int n) { int difference = 0, ans = 0, count = k; bool flag = false; // interating over the array for(int i = 0 ; i < n - 1; i++) { difference = 0; // check if i-th and // (i + 1)-th element // are not consecutive if ((a[i] + 1) != a[i + 1]) { // save their difference difference += (a[i + 1] - a[i]) - 1; // check for difference // and given k if (difference >= count) { ans = a[i] + count; flag = true; break; } else count -= difference; } } // if found if(flag) return ans; else return -1; } // Driver code public static void Main() { // Input array int []a = {1, 5, 11, 19}; // k-th missing element // to be found in the array int k = 11; int n = a.Length; // calling function to // find missing element int missing = missingK(a, k, n); Console.Write(missing); } } // This code is contributed by // Manish Shaw (manishshaw1) |
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PHP
<?php // Function to find k-th // missing element function missingK(&$a, $k, $n) { $difference = 0; $ans = 0; $count = $k; $flag = 0; // interating over the array for($i = 0 ; $i < $n - 1; $i++) { $difference = 0; // check if i-th and // (i + 1)-th element // are not consecutive if (($a[$i] + 1) != $a[$i + 1]) { // save their difference $difference += ($a[$i + 1] - $a[$i]) - 1; // check for difference // and given k if ($difference >= $count) { $ans = $a[$i] + $count; $flag = 1; break; } else $count -= $difference; } } // if found if($flag) return $ans; else return -1; } // Driver Code // Input array $a = array(1, 5, 11, 19); // k-th missing element // to be found in the array $k = 11; $n = count($a); // calling function to // find missing element $missing = missingK($a, $k, $n); echo $missing; // This code is contributed by Manish Shaw // (manishshaw1) ?> |
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Output :
14
Time Complexity :O(n), where n is the number of elements in the array.
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Improved By : manishshaw1
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