Given an array arr[] of positive integers, the task is to find whether it is possible to make this array strictly decreasing by modifying at most one element.
Examples:
Input: arr[] = {12, 9, 10, 5, 2}
Output: Yes
{12, 11, 10, 5, 2} is one of the valid solutions.
Input: arr[] = {1, 2, 3, 4}
Output: No
Approach: For every element arr[i], if it is greater than both arr[i – 1] and arr[i + 1] or it is smaller than both arr[i – 1] and arr[i + 1] then arr[i] needs to be modified.
i.e arr[i] = (arr[i – 1] + arr[i + 1]) / 2. If after modification, arr[i] = arr[i – 1] or arr[i + 1] then the array cannot be made strictly decreasing without affecting at most one element else count all such modifications, if the count of modifications in the end is less than or equal to 1 then print Yes else print No.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function that returns true if the array // can be made strictly decreasing // with at most one change bool check(vector< int > arr, int n)
{ // To store the number of modifications
// required to make the array
// strictly decreasing
int modify = 0;
// Check whether the last element needs
// to be modify or not
if (arr[n - 1] >= arr[n - 2]) {
arr[n - 1] = arr[n - 2] - 1;
modify++;
}
// Check whether the first element needs
// to be modify or not
if (arr[0] <= arr[1]) {
arr[0] = arr[1] + 1;
modify++;
}
// Loop from 2nd element to the 2nd last element
for ( int i = n - 2; i > 0; i--) {
// Check whether arr[i] needs to be modified
if ((arr[i - 1] <= arr[i] && arr[i + 1] <= arr[i])
|| (arr[i - 1] >= arr[i] && arr[i + 1] >= arr[i])) {
// Modifying arr[i]
arr[i] = (arr[i - 1] + arr[i + 1]) / 2;
modify++;
// Check if arr[i] is equal to any of
// arr[i-1] or arr[i+1]
if (arr[i] == arr[i - 1] || arr[i] == arr[i + 1])
return false ;
}
}
// If more than 1 modification is required
if (modify > 1)
return false ;
return true ;
} // Driver code int main()
{ vector< int > arr = { 10, 5, 11, 2 };
int n = arr.size();
if (check(arr, n))
cout << "Yes" ;
else
cout << "No" ;
return 0;
} |
// Java implementation of the approach class GFG {
// Function that returns true if the array
// can be made strictly decreasing
// with at most one change
public static boolean check( int [] arr, int n)
{
// To store the number of modifications
// required to make the array
// strictly decreasing
int modify = 0 ;
// Check whether the last element needs
// to be modify or not
if (arr[n - 1 ] >= arr[n - 2 ]) {
arr[n - 1 ] = arr[n - 2 ] - 1 ;
modify++;
}
// Check whether the first element needs
// to be modify or not
if (arr[ 0 ] <= arr[ 1 ]) {
arr[ 0 ] = arr[ 1 ] + 1 ;
modify++;
}
// Loop from 2nd element to the 2nd last element
for ( int i = n - 2 ; i > 0 ; i--) {
// Check whether arr[i] needs to be modified
if ((arr[i - 1 ] <= arr[i] && arr[i + 1 ] <= arr[i])
|| (arr[i - 1 ] >= arr[i] && arr[i + 1 ] >= arr[i])) {
// Modifying arr[i]
arr[i] = (arr[i - 1 ] + arr[i + 1 ]) / 2 ;
modify++;
// Check if arr[i] is equal to any of
// arr[i-1] or arr[i+1]
if (arr[i] == arr[i - 1 ] || arr[i] == arr[i + 1 ])
return false ;
}
}
// If more than 1 modification is required
if (modify > 1 )
return false ;
return true ;
}
// Driver code
public static void main(String[] args)
{
int [] arr = { 10 , 5 , 11 , 3 };
int n = arr.length;
if (check(arr, n))
System.out.print( "Yes" );
else
System.out.print( "No" );
}
} |
# Python3 implementation of the approach # Function that returns true if the array # can be made strictly decreasing # with at most one change def check(arr, n):
modify = 0
# Check whether the last element needs
# to be modify or not
if (arr[n - 1 ] > = arr[n - 2 ]):
arr[n - 1 ] = arr[n - 2 ] - 1
modify + = 1
# Check whether the first element needs
# to be modify or not
if (arr[ 0 ] < = arr[ 1 ]):
arr[ 0 ] = arr[ 1 ] + 1
modify + = 1
# Loop from 2nd element to the 2nd last element
for i in range (n - 2 , 0 , - 1 ):
# Check whether arr[i] needs to be modified
if (arr[i - 1 ] < = arr[i] and arr[i + 1 ] < = arr[i]) or \
(arr[i - 1 ] > = arr[i] and arr[i + 1 ] > = arr[i]):
# Modifying arr[i]
arr[i] = (arr[i - 1 ] + arr[i + 1 ]) / / 2
modify + = 1
# Check if arr[i] is equal to any of
# arr[i-1] or arr[i + 1]
if (arr[i] = = arr[i - 1 ] or arr[i] = = arr[i + 1 ]):
return False
# If more than 1 modification is required
if (modify > 1 ):
return False
return True
# Driver code if __name__ = = "__main__" :
arr = [ 10 , 5 , 11 , 3 ]
n = len (arr)
if (check(arr, n)):
print ( "Yes" )
else :
print ( "No" )
|
// C# implementation of the approach using System;
class GFG
{ // Function that returns true if the array
// can be made strictly decreasing
// with at most one change
public static bool check( int []arr, int n)
{
// To store the number of modifications
// required to make the array
// strictly decreasing
int modify = 0;
// Check whether the last element needs
// to be modify or not
if (arr[n - 1] >= arr[n - 2])
{
arr[n - 1] = arr[n - 2] - 1;
modify++;
}
// Check whether the first element needs
// to be modify or not
if (arr[0] <= arr[1])
{
arr[0] = arr[1] + 1;
modify++;
}
// Loop from 2nd element to the 2nd last element
for ( int i = n - 2; i > 0; i--)
{
// Check whether arr[i] needs to be modified
if ((arr[i - 1] <= arr[i] && arr[i + 1] <= arr[i])
|| (arr[i - 1] >= arr[i] && arr[i + 1] >= arr[i]))
{
// Modifying arr[i]
arr[i] = (arr[i - 1] + arr[i + 1]) / 2;
modify++;
// Check if arr[i] is equal to any of
// arr[i-1] or arr[i+1]
if (arr[i] == arr[i - 1] || arr[i] == arr[i + 1])
return false ;
}
}
// If more than 1 modification is required
if (modify > 1)
return false ;
return true ;
}
// Driver code
static public void Main ()
{
int []arr = { 10, 5, 11, 3 };
int n = arr.Length;
if (check(arr, n))
Console.Write( "Yes" );
else
Console.Write( "No" );
}
} // This code is contributed by ajit. |
<script> // Javascript implementation of the approach
// Function that returns true if the array
// can be made strictly decreasing
// with at most one change
function check(arr, n)
{
// To store the number of modifications
// required to make the array
// strictly decreasing
let modify = 0;
// Check whether the last element needs
// to be modify or not
if (arr[n - 1] >= arr[n - 2])
{
arr[n - 1] = arr[n - 2] - 1;
modify++;
}
// Check whether the first element needs
// to be modify or not
if (arr[0] <= arr[1])
{
arr[0] = arr[1] + 1;
modify++;
}
// Loop from 2nd element to the 2nd last element
for (let i = n - 2; i > 0; i--)
{
// Check whether arr[i] needs to be modified
if ((arr[i - 1] <= arr[i] && arr[i + 1] <= arr[i])
|| (arr[i - 1] >= arr[i] && arr[i + 1] >= arr[i]))
{
// Modifying arr[i]
arr[i] = parseInt((arr[i - 1] + arr[i + 1]) / 2, 10);
modify++;
// Check if arr[i] is equal to any of
// arr[i-1] or arr[i+1]
if (arr[i] == arr[i - 1] || arr[i] == arr[i + 1])
return false ;
}
}
// If more than 1 modification is required
if (modify > 1)
return false ;
return true ;
}
let arr = [ 10, 5, 11, 3 ];
let n = arr.length;
if (check(arr, n))
document.write( "Yes" );
else
document.write( "No" );
</script> |
Yes
Time Complexity: O(N)
Auxiliary Space: O(1)