Given an array arr[0..n-1] of distinct elements and a range [low, high], find all numbers that are in range, but not in array. The missing elements should be printed in sorted order.
Examples:
Input: arr[] = {10, 12, 11, 15}, low = 10, hight = 15 Output: 13, 14 Input: arr[] = {1, 14, 11, 51, 15}, low = 50, hight = 55 Output: 50, 52, 53, 54
There can be following two approaches to solve the problem.
- Use Sorting : Sort the array, then do binary search for ‘low’. Once location of low is find, start traversing array from that location and keep printing all missing numbers.
C++
// A sorting based C++ program to find missing
// elements from an array
#include<bits/stdc++.h>
using
namespace
std;
// Print all elements of range [low, high] that
// are not present in arr[0..n-1]
void
printMissing(
int
arr[],
int
n,
int
low,
int
high)
{
// Sort the array
sort(arr, arr+n);
// Do binary search for 'low' in sorted
// array and find index of first element
// which either equal to or greater than
// low.
int
*ptr = lower_bound(arr, arr+n, low);
int
index = ptr - arr;
// Start from the found index and linearly
// search every range element x after this
// index in arr[]
int
i = index, x = low;
while
(i < n && x<=high)
{
// If x doesn't math with current element
// print it
if
(arr[i] != x)
cout << x <<
" "
;
// If x matches, move to next element in arr[]
else
i++;
// Move to next element in range [low, high]
x++;
}
// Print range elements thar are greater than the
// last element of sorted array.
while
(x <= high)
cout << x++ <<
" "
;
}
// Driver program
int
main()
{
int
arr[] = {1, 3, 5, 4};
int
n =
sizeof
(arr)/
sizeof
(arr[0]);
int
low = 1, high = 10;
printMissing(arr, n, low, high);
return
0;
}
Java
// A sorting based Java program to find missing
// elements from an array
import
java.util.Arrays;
public
class
PrintMissing
{
// Print all elements of range [low, high] that
// are not present in arr[0..n-1]
static
void
printMissing(
int
ar[],
int
low,
int
high)
{
Arrays.sort(ar);
// Do binary search for 'low' in sorted
// array and find index of first element
// which either equal to or greater than
// low.
int
index = ceilindex(ar, low,
0
, ar.length -
1
);
int
x = low;
// Start from the found index and linearly
// search every range element x after this
// index in arr[]
while
(index < ar.length && x <= high)
{
// If x doesn't math with current element
// print it
if
(ar[index] != x)
{
System.out.print(x +
" "
);
}
// If x matches, move to next element in arr[]
else
index++;
// Move to next element in range [low, high]
x++;
}
// Print range elements thar are greater than the
// last element of sorted array.
while
(x <= high)
{
System.out.print(x +
" "
);
x++;
}
}
// Utility function to find ceil index of given element
static
int
ceilindex(
int
ar[],
int
val,
int
low,
int
high)
{
if
(val < ar[
0
])
return
0
;
if
(val > ar[ar.length -
1
])
return
ar.length;
int
mid = (low + high) /
2
;
if
(ar[mid] == val)
return
mid;
if
(ar[mid] < val)
{
if
(mid +
1
< high && ar[mid +
1
] >= val)
return
mid +
1
;
return
ceilindex(ar, val, mid +
1
, high);
}
else
{
if
(mid -
1
>= low && ar[mid -
1
] < val)
return
mid;
return
ceilindex(ar, val, low, mid -
1
);
}
}
// Driver program to test above function
public
static
void
main(String[] args)
{
int
arr[] = {
1
,
3
,
5
,
4
};
int
low =
1
, high =
10
;
printMissing(arr, low, high);
}
}
// This code is contributed by Rishabh Mahrsee
C#
// A sorting based Java program to
// find missing elements from an array
using
System;
class
GFG
{
// Print all elements of range
// [low, high] that are not
// present in arr[0..n-1]
static
void
printMissing(
int
[]ar,
int
low,
int
high)
{
Array.Sort(ar);
// Do binary search for 'low' in sorted
// array and find index of first element
// which either equal to or greater than
// low.
int
index = ceilindex(ar, low, 0,
ar.Length - 1);
int
x = low;
// Start from the found index and linearly
// search every range element x after this
// index in arr[]
while
(index < ar.Length && x <= high)
{
// If x doesn't math with current
// element print it
if
(ar[index] != x)
{
Console.Write(x +
" "
);
}
// If x matches, move to next
// element in arr[]
else
index++;
// Move to next element in
// range [low, high]
x++;
}
// Print range elements thar
// are greater than the
// last element of sorted array.
while
(x <= high)
{
Console.Write(x +
" "
);
x++;
}
}
// Utility function to find
// ceil index of given element
static
int
ceilindex(
int
[]ar,
int
val,
int
low,
int
high)
{
if
(val < ar[0])
return
0;
if
(val > ar[ar.Length - 1])
return
ar.Length;
int
mid = (low + high) / 2;
if
(ar[mid] == val)
return
mid;
if
(ar[mid] < val)
{
if
(mid + 1 < high && ar[mid + 1] >= val)
return
mid + 1;
return
ceilindex(ar, val, mid + 1, high);
}
else
{
if
(mid - 1 >= low && ar[mid - 1] < val)
return
mid;
return
ceilindex(ar, val, low, mid - 1);
}
}
// Driver Code
static
public
void
Main ()
{
int
[]arr = { 1, 3, 5, 4 };
int
low = 1, high = 10;
printMissing(arr, low, high);
}
}
// This code is contributed
// by Sach_Code
Output:2 6 7 8 9 10
- Use Hashing : Create a hash table and insert all array items into the hash table. Once all items are in hash table, traverse through the range and print all missing elements.
C++
// A hashing based C++ program to find missing
// elements from an array
#include<bits/stdc++.h>
using
namespace
std;
// Print all elements of range [low, high] that
// are not present in arr[0..n-1]
void
printMissing(
int
arr[],
int
n,
int
low,
int
high)
{
// Insert all elements of arr[] in set
unordered_set<
int
> s;
for
(
int
i=0; i<n; i++)
s.insert(arr[i]);
// Traverse throught the range an print all
// missing elements
for
(
int
x=low; x<=high; x++)
if
(s.find(x) == s.end())
cout << x <<
" "
;
}
// Driver program
int
main()
{
int
arr[] = {1, 3, 5, 4};
int
n =
sizeof
(arr)/
sizeof
(arr[0]);
int
low = 1, high = 10;
printMissing(arr, n, low, high);
return
0;
}
Java
// A hashing based Java program to find missing
// elements from an array
import
java.util.Arrays;
import
java.util.HashSet;
public
class
Print
{
// Print all elements of range [low, high] that
// are not present in arr[0..n-1]
static
void
printMissing(
int
ar[],
int
low,
int
high)
{
HashSet<Integer> hs =
new
HashSet<>();
// Insert all elements of arr[] in set
for
(
int
i =
0
; i < ar.length; i++)
hs.add(ar[i]);
// Traverse throught the range an print all
// missing elements
for
(
int
i = low; i <= high; i++)
{
if
(!hs.contains(i))
{
System.out.print(i +
" "
);
}
}
}
// Driver program to test above function
public
static
void
main(String[] args)
{
int
arr[] = {
1
,
3
,
5
,
4
};
int
low =
1
, high =
10
;
printMissing(arr, low, high);
}
}
// This code is contributed by Rishabh Mahrsee
Python3
# A hashing based Python 3 program to
# find missing elements from an array
# Print all elements of range
# [low, high] that are not
# present in arr[0..n-1]
def
printMissing(arr, n, low, high):
# Insert all elements of
# arr[] in set
s
=
set
(arr)
# Traverse through the range
# and print all missing elements
for
x
in
range
(low, high
+
1
):
if
x
not
in
s:
print
(x, end
=
' '
)
# Driver Code
arr
=
[
1
,
3
,
5
,
4
]
n
=
len
(arr)
low, high
=
1
,
10
printMissing(arr, n, low, high)
# This code is contributed
# by SamyuktaSHegde
C#
// A hashing based C# program to
// find missing elements from an array
using
System;
using
System.Collections.Generic;
class
GFG
{
// Print all elements of range
// [low, high] that are not
// present in arr[0..n-1]
static
void
printMissing(
int
[]arr,
int
n,
int
low,
int
high)
{
// Insert all elements of arr[] in set
HashSet<
int
> s =
new
HashSet<
int
>();
for
(
int
i = 0; i < n; i++)
{
s.Add(arr[i]);
}
// Traverse throught the range
// an print all missing elements
for
(
int
x = low; x <= high; x++)
if
(!s.Contains(x))
Console.Write(x +
" "
);
}
// Driver Code
public
static
void
Main()
{
int
[]arr = {1, 3, 5, 4};
int
n = arr.Length;
int
low = 1, high = 10;
printMissing(arr, n, low, high);
}
}
// This code is contributed by ihritik
Output:2 6 7 8 9 10
Which approach is better?
Time complexity of first approach is O(nLogn + k) where k is number of missing elements (Note that k may be more than nLogn if array is small and range is big)
Time complexity of second solution is O(n + (high-low+1)).
If the given array has almost elements of range i.e., n is close to value of (high-low+1), then second approach is definitely better as there is no Log n factor. But if n is much smaller than range, then first approach is better as it doesn’t require extra space for hashing. We can also modify first approach to print adjacent missing elements as range to save time. For example if 50, 51, 52, 53, 54, 59 are missing, we can print them as 50-54, 59 in first method. And if printing this way is allowed, the first approach takes only O(n Log n) time.
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