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 : 1 Explanation: Missing Element in the increasing sequence are {1,4, 6, 7, 8}. So k-th missing element is 1 Input : a[] = {2, 3, 5, 9, 10, 11, 12}; k = 4; Output : 7 Explanation: missing element in the increasing sequence are {1, 4, 6, 7, 8} so k-th missing element is 7
Approach 1: Start iterating over the array elements, and for every element check if the next element is consecutive or not, if not, then take the difference between these two, and check if the difference is greater than or equal to given k, then calculate ans = a[i] + count, else iterate for next element.
Implementation:
#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;
// case when first number is not 1
if (a[0] != 1) {
difference = a[0] - 1;
if (difference >= count)
return count;
count -= difference;
}
// iterating 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;
} |
// 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 ;
// case when first number is not 1
if (a[ 0 ] != 1 ) {
difference = a[ 0 ] - 1 ;
if (difference >= count)
return count;
count -= difference;
}
// iterating 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) |
# Function to find k-th # missing element def missingK(a, k, n):
difference = 0
ans = 0
count = k
flag = 0
# case when first number is not 1
if a[ 0 ] ! = 1 :
difference = a[ 0 ] - 1
if difference > = count:
return count
count - = difference
# iterating 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 = 11
n = len (a)
# calling function to # find missing element missing = missingK(a, k, n)
print (missing)
# This code is contributed by # Manish Shaw (manishshaw1) |
// 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 ;
// case when first number is not 1
if (a[0] != 1) {
difference = a[0] - 1;
if (difference >= count)
return count;
count -= difference;
}
// iterating 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) |
<?php // Function to find k-th // missing element function missingK(& $a , $k , $n )
{ $difference = 0;
$ans = 0;
$count = $k ;
$flag = 0;
// iterating 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) ?> |
<script> // Javascript program to check for // even or odd // Function to find k-th // missing element function missingK(a, k, n)
{ let difference = 0, ans = 0, count = k;
let flag = false ;
//case when first number is not 1
if (a[0] != 1){
difference = a[0]-1;
if (difference >= count){
return count;
}
count -= difference;
}
// iterating over the array
for (let 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 // Input array let a = [ 1, 5, 11, 19 ]; // k-th missing element // to be found in the array let k = 11; let n = a.length; // Calling function to // find missing element let missing = missingK(a, k, n); document.write(missing); // This code is contributed by suresh07 </script> |
14
Time Complexity: O(n), where n is the number of elements in the array.
Auxiliary Space: O(1)
Approach 2:
Apply a binary search. Since the array is sorted we can find at any given index how many numbers are missing as arr[index] – (index+1). We would leverage this knowledge and apply binary search to narrow down our hunt to find that index from which getting the missing number is easier.
Implementation:
// CPP program for above approach #include <bits/stdc++.h> #include <iostream> using namespace std;
// Function to find // kth missing number int missingK(vector< int >& arr, int k)
{ int n = arr.size();
int l = 0, u = n - 1, mid;
while (l <= u) {
mid = (l + u) / 2;
int numbers_less_than_mid = arr[mid] - (mid + 1);
// If the total missing number
// count is equal to k we can iterate
// backwards for the first missing number
// and that will be the answer.
if (numbers_less_than_mid == k) {
// To further optimize we check
// if the previous element's
// missing number count is equal
// to k. Eg: arr = [4,5,6,7,8]
// If you observe in the example array,
// the total count of missing numbers for all
// the indices are same, and we are
// aiming to narrow down the
// search window and achieve O(logn)
// time complexity which
// otherwise would've been O(n).
if (mid > 0 && (arr[mid - 1] - (mid)) == k) {
u = mid - 1;
continue ;
}
// Else we return arr[mid] - 1.
return arr[mid] - 1;
}
// Here we appropriately
// narrow down the search window.
if (numbers_less_than_mid < k) {
l = mid + 1;
}
else if (k < numbers_less_than_mid) {
u = mid - 1;
}
}
// In case the upper limit is -ve
// it means the missing number set
// is 1,2,..,k and hence we directly return k.
if (u < 0)
return k;
// Else we find the residual count
// of numbers which we'd then add to
// arr[u] and get the missing kth number.
int less = arr[u] - (u + 1);
k -= less;
// Return arr[u] + k
return arr[u] + k;
} // Driver Code int main()
{ vector< int > arr = { 2, 3, 4, 7, 11 };
int k = 5;
// Function Call
cout << "Missing kth number = " << missingK(arr, k)
<< endl;
return 0;
} |
// Java program for above approach public class GFG {
// Function to find
// kth missing number
static int missingK( int [] arr, int k)
{
int n = arr.length;
int l = 0 , u = n - 1 , mid;
while (l <= u) {
mid = (l + u) / 2 ;
int numbers_less_than_mid
= arr[mid] - (mid + 1 );
// If the total missing number
// count is equal to k we can iterate
// backwards for the first missing number
// and that will be the answer.
if (numbers_less_than_mid == k) {
// To further optimize we check
// if the previous element's
// missing number count is equal
// to k. Eg: arr = [4,5,6,7,8]
// If you observe in the example array,
// the total count of missing numbers for
// all the indices are same, and we are
// aiming to narrow down the
// search window and achieve O(logn)
// time complexity which
// otherwise would've been O(n).
if (mid > 0
&& (arr[mid - 1 ] - (mid)) == k) {
u = mid - 1 ;
continue ;
}
// Else we return arr[mid] - 1.
return arr[mid] - 1 ;
}
// Here we appropriately
// narrow down the search window.
if (numbers_less_than_mid < k) {
l = mid + 1 ;
}
else if (k < numbers_less_than_mid) {
u = mid - 1 ;
}
}
// In case the upper limit is -ve
// it means the missing number set
// is 1,2,..,k and hence we directly return k.
if (u < 0 )
return k;
// Else we find the residual count
// of numbers which we'd then add to
// arr[u] and get the missing kth number.
int less = arr[u] - (u + 1 );
k -= less;
// Return arr[u] + k
return arr[u] + k;
}
// Driver code
public static void main(String[] args)
{
int [] arr = { 2 , 3 , 4 , 7 , 11 };
int k = 5 ;
// Function Call
System.out.println( "Missing kth number = "
+ missingK(arr, k));
}
} // This code is contributed by divyesh072019. |
# Python3 program for above approach # Function to find # kth missing number def missingK(arr, k):
n = len (arr)
l = 0
u = n - 1
mid = 0
while (l < = u):
mid = (l + u) / / 2
numbers_less_than_mid = arr[mid] - (mid + 1 )
# If the total missing number
# count is equal to k we can iterate
# backwards for the first missing number
# and that will be the answer.
if (numbers_less_than_mid = = k):
# To further optimize we check
# if the previous element's
# missing number count is equal
# to k. Eg: arr = [4,5,6,7,8]
# If you observe in the example array,
# the total count of missing numbers for all
# the indices are same, and we are
# aiming to narrow down the
# search window and achieve O(logn)
# time complexity which
# otherwise would've been O(n).
if (mid > 0 and (arr[mid - 1 ] - (mid)) = = k):
u = mid - 1
continue
# Else we return arr[mid] - 1.
return arr[mid] - 1
# Here we appropriately
# narrow down the search window.
if (numbers_less_than_mid < k):
l = mid + 1
elif (k < numbers_less_than_mid):
u = mid - 1
# In case the upper limit is -ve
# it means the missing number set
# is 1,2,..,k and hence we directly return k.
if (u < 0 ):
return k
# Else we find the residual count
# of numbers which we'd then add to
# arr[u] and get the missing kth number.
less = arr[u] - (u + 1 )
k - = less
# Return arr[u] + k
return arr[u] + k
# Driver Code if __name__ = = '__main__' :
arr = [ 2 , 3 , 4 , 7 , 11 ]
k = 5
# Function Call
print ( "Missing kth number = " + str (missingK(arr, k)))
# This code is contributed by rutvik_56. |
// C# program for above approach using System;
class GFG {
// Function to find
// kth missing number
static int missingK( int [] arr, int k)
{
int n = arr.Length;
int l = 0, u = n - 1, mid;
while (l <= u) {
mid = (l + u) / 2;
int numbers_less_than_mid
= arr[mid] - (mid + 1);
// If the total missing number
// count is equal to k we can iterate
// backwards for the first missing number
// and that will be the answer.
if (numbers_less_than_mid == k) {
// To further optimize we check
// if the previous element's
// missing number count is equal
// to k. Eg: arr = [4,5,6,7,8]
// If you observe in the example array,
// the total count of missing numbers for
// all the indices are same, and we are
// aiming to narrow down the
// search window and achieve O(logn)
// time complexity which
// otherwise would've been O(n).
if (mid > 0
&& (arr[mid - 1] - (mid)) == k) {
u = mid - 1;
continue ;
}
// Else we return arr[mid] - 1.
return arr[mid] - 1;
}
// Here we appropriately
// narrow down the search window.
if (numbers_less_than_mid < k) {
l = mid + 1;
}
else if (k < numbers_less_than_mid) {
u = mid - 1;
}
}
// In case the upper limit is -ve
// it means the missing number set
// is 1,2,..,k and hence we directly return k.
if (u < 0)
return k;
// Else we find the residual count
// of numbers which we'd then add to
// arr[u] and get the missing kth number.
int less = arr[u] - (u + 1);
k -= less;
// Return arr[u] + k
return arr[u] + k;
}
// Driver code
static void Main()
{
int [] arr = { 2, 3, 4, 7, 11 };
int k = 5;
// Function Call
Console.WriteLine( "Missing kth number = "
+ missingK(arr, k));
}
} // This code is contributed by divyeshrabadiya07. |
<script> // JavaScript program for above approach // Function to find
// kth missing number
function missingK(arr, k)
{
var n = arr.length;
var l = 0, u = n - 1, mid;
while (l <= u)
{
mid = (l + u)/2;
var numbers_less_than_mid = arr[mid] -
(mid + 1);
// If the total missing number
// count is equal to k we can iterate
// backwards for the first missing number
// and that will be the answer.
if (numbers_less_than_mid == k)
{
// To further optimize we check
// if the previous element's
// missing number count is equal
// to k. Eg: arr = [4,5,6,7,8]
// If you observe in the example array,
// the total count of missing numbers for all
// the indices are same, and we are
// aiming to narrow down the
// search window and achieve O(logn)
// time complexity which
// otherwise would've been O(n).
if (mid > 0 && (arr[mid - 1] - (mid)) == k)
{
u = mid - 1;
continue ;
}
// Else we return arr[mid] - 1.
return arr[mid] - 1;
}
// Here we appropriately
// narrow down the search window.
if (numbers_less_than_mid < k)
{
l = mid + 1;
}
else if (k < numbers_less_than_mid)
{
u = mid - 1;
}
}
// In case the upper limit is -ve
// it means the missing number set
// is 1,2,..,k and hence we directly return k.
if (u < 0)
return k;
// Else we find the residual count
// of numbers which we'd then add to
// arr[u] and get the missing kth number.
var less = arr[u] - (u + 1);
k -= less;
// Return arr[u] + k
return arr[u] + k;
}
// Driver code
var arr = [2,3,4,7,11];
var k = 5;
// Function Call
document.write( "Missing kth number = " + missingK(arr, k));
// This code is contributed by shivanisinghss2110 </script> |
Missing kth number = 9
Time Complexity: O(logn), where n is the number of elements in the array.
Auxiliary Space: O(1)
Approach 3 (Using Map):
We will traverse the array and mark each of the elements as visited in the map and we will also keep track of the min and max element present so that we know the lower and upper bound for the given particular input. Then we start a loop from lower to upper bound and maintain a count variable. As soon we found an element that is not present in the map we increment the count and until the count becomes equal to k.
Implementation:
#include <iostream> #include <unordered_map> using namespace std;
int solve( int arr[], int k, int n)
{ unordered_map< int , int > umap;
int mins = 99999;
int maxs = -99999;
for ( int i = 0; i < n; i++) {
umap[arr[i]]
= 1; // mark each element of array in map
if (mins > arr[i])
mins = arr[i]; // keeping track of minimum
// element
if (maxs < arr[i])
maxs = arr[i]; // keeping track of maximum
// element i.e. upper bound
}
int counts = 0;
// iterate from lower to upper bound
for ( int i = mins; i <= maxs; i++) {
if (umap[i] == 0)
counts++;
if (counts == k)
return i;
}
return -1;
} int main()
{ int arr[] = { 2, 3, 5, 9, 10, 11, 12 };
int k = 4;
cout << solve(arr, k, 7); //(array , k , size of array)
return 0;
} |
// Java approach // Importing HashMap class import java.util.HashMap;
class GFG {
public static int solve( int arr[], int k, int n)
{
HashMap<Integer, Integer> umap
= new HashMap<Integer, Integer>();
int mins = 99999 ;
int maxs = - 99999 ;
for ( int i = 0 ; i < n; i++) {
umap.put(
arr[i],
1 ); // mark each element of array in map
if (mins > arr[i])
mins = arr[i]; // keeping track of minimum
// element
if (maxs < arr[i])
maxs = arr[i]; // keeping track of maximum
// element i.e. upper bound
}
int counts = 0 ;
// iterate from lower to upper bound
for ( int i = mins; i <= maxs; i++) {
if (umap.get(i) == null )
counts++;
if (counts == k)
return i;
}
return - 1 ;
}
public static void main(String[] args)
{
int arr[] = { 2 , 3 , 5 , 9 , 10 , 11 , 12 };
int k = 4 ;
System.out.println(
solve(arr, k, 7 )); //(array , k , size of array)
}
} // This code is contributed by Shubham Singh |
# Python approach def solve(arr, k, n):
umap = {}
mins = 99999
maxs = - 99999
for i in range (n):
umap[arr[i]] = 1 # mark each element of array in map
if (mins > arr[i]):
mins = arr[i] # keeping track of minimum element
if (maxs < arr[i]):
maxs = arr[i] # keeping track of maximum element i.e. upper bound
counts = 0
# iterate from lower to upper bound
for i in range (mins, maxs + 1 ):
if (i not in umap):
counts + = 1
if (counts = = k):
return i
return - 1
arr = [ 2 , 3 , 5 , 9 , 10 , 11 , 12 ]
k = 4
print (solve(arr, k, 7 )) # (array , k , size of array)
# This code is contributed # by Shubham Singh |
// C# program for above approach using System;
using System.Collections.Generic;
class GFG {
// Function to find
// kth missing number
static int solve( int [] arr, int k, int n)
{
Dictionary< int , int > umap
= new Dictionary< int , int >();
int mins = 99999;
int maxs = -99999;
for ( int i = 0; i < n; i++) {
umap.Add(
arr[i],
1); // mark each element of array in map
if (mins > arr[i])
mins = arr[i]; // keeping track of minimum
// element
if (maxs < arr[i])
maxs = arr[i]; // keeping track of maximum
// element i.e. upper bound
}
int counts = 0;
// iterate from lower to upper bound
for ( int i = mins; i <= maxs; i++) {
if (!umap.ContainsKey(i))
counts++;
if (counts == k)
return i;
}
return -1;
}
// Driver code
static void Main()
{
int [] arr = { 2, 3, 5, 9, 10, 11, 12 };
int k = 4;
int n = arr.Length;
// Function Call
Console.WriteLine( "Missing kth number = "
+ solve(arr, k, n));
//(array , k , size of array));
}
} // This code is contributed by Aarti_Rathi |
<script> // JavaScript program function solve(arr ,k ,n){
let umap = new Map()
let mins = 99999
let maxs = -99999
for (let i = 0; i < n; i++){
umap.set(arr[i] , 1) //mark each element of array in map
if (mins > arr[i])
mins = arr[i] //keeping track of minimum element
if (maxs < arr[i])
maxs = arr[i] //keeping track of maximum element i.e. upper bound
}
let counts = 0
// iterate from lower to upper bound
for (let i = mins; i < maxs + 1; i++){
if (!umap.has(i))
counts += 1
if (counts == k)
return i
}
return -1
} // driver code let arr = [2, 3, 5, 9, 10, 11, 12] let k = 4 document.write(solve(arr , k , 7), "</br>" ) //(array , k , size of array)
// This code is contributed by shinjanpatra </script> |
8
Time Complexity: O(n+m), where n is the number of elements in the array and m is the difference between the largest and smallest element of the array.
Auxiliary Space: O(n)