Replacing an element makes array elements consecutive
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++
#include <bits/stdc++.h>
using namespace std;
int findElement( int arr[], int n)
{
sort(arr, arr+n);
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 (mismatch_count1 == 1)
return res;
if (mismatch_count1 == 0)
return 0;
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 (mismatch_count2 == 1)
return res;
return -1;
}
int 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;
}
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Java
import java.io.*;
import java.util.Arrays;
class GFG
{
static int findElement( int []arr,
int n)
{
Arrays.sort(arr);
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 (mismatch_count1 == 1 )
return res;
if (mismatch_count1 == 0 )
return 0 ;
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 (mismatch_count2 == 1 )
return res;
return - 1 ;
}
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);
}
}
|
Python3
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()
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 (mismatch_count1 = = 1 ):
return res
if (mismatch_count1 = = 0 ):
return 0
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 (mismatch_count2 = = 1 ):
return res
return - 1
if __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)
|
C#
using System;
using System.Linq;
using System.Collections.Generic;
class GFG
{
static int findElement( int []arr,
int n)
{
Array.Sort(arr);
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 (mismatch_count1 == 1)
return res;
if (mismatch_count1 == 0)
return 0;
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 (mismatch_count2 == 1)
return res;
return -1;
}
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);
}
}
|
Javascript
<script>
function binary_search(arr, x) {
let start=0, end=arr.length-1;
while (start<=end){
let mid=Math.floor((start + end)/2);
if (arr[mid]===x) return mid;
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});
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 (mismatch_count1 == 1)
return res;
if (mismatch_count1 == 0)
return 0;
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 (mismatch_count2 == 1)
return res;
return -1;
}
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>
|
Time Complexity: O(n Log n)
Auxiliary Space: O(1)
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