Hello Everyone,
Given an array Arr of positive integers and a value X. The task is to find the number of values that is greater than or equal to X.
But the twist is that the values of the array are kept changing after every operation. There are two possibilities:
If the current value is picked then all the remaining values in the array will be decreased by 1.
If the current value is not picked then all the remaining values in the array will be increased by 1.
Examples:
Input: arr[] = {10, 5, 5, 4, 9}, X = 10
Output: 2
Explanation:
Arr = {10, 5, 5, 4, 9}, pos = 0
10 is picked
Arr = {10, 4, 4, 3, 8}, pos = 1
4 is not picked
Arr = {10, 4, 5, 4, 9}, pos = 2
5 is not picked
Arr = {10, 4, 5, 5, 10}, pos = 3
5 is not picked
Arr = {10, 4, 5, 5, 11}, pos = 4
11 is picked
Hence two elements are picked.
Input: arr[] = {5, 4, 3, 2, 1}, X = 4
Output: 1
Recommended: Please try your approach on {IDE} first, before moving on to the solution.
Naive Approach: The idea is to iterate through every value in an array and check whether the ith value is greater, lesser, or equal than required value X. If the ith value is less than required value then increase the value from (i+1)th to end of the array by 1 else If the ith value is greater or equal than required value X then decrease the value by 1 from (i+1)th to end of the array.
Below is the implementation of the above approach:
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to count number
// of values greater or equal to x
int increaseDecreaseValue(int arr[],
int x, int n)
{
int TotalValue = 0;
for (int i = 0; i < n; i++)
{
if (arr[i] < x)
{
// Current value is less
// than required value
// then we need to increase
// the value from i + 1 to
// len(arr) by 1
for (int j = i + 1; j < n; j++)
{
arr[j] += 1;
}
}
else
{
// Current value is greater
// or equal to required
// value then we need to
// decrease the value from
// (i + 1) to len(arr)-1 by 1
TotalValue += 1;
for (int j = i + 1; j < n; j++)
{
if (arr[j] == 0)
{
continue;
}
else
{
arr[j] -= 1;
}
}
}
}
return TotalValue;
}
// Driver Code
int main()
{
int x = 4;
int arr[] = {5, 4, 3, 2, 1};
int n = sizeof(arr) / sizeof(arr[0]);
int countValue = increaseDecreaseValue(arr, x, n);
cout << countValue;
return 0;
}
// This code is contributed by Rajput-Ji
Output:
1
ime Complexity: O(N^{2})
Efficient Approach:
This problem can further be optimized to O(N) .
Here the main idea is to check by how much this index value should change.
This can be done by using a temporary variable, here it is currentStatus that will keep the net effect on the current index by the previous decisions.
The effect will be added to the value of that index and that will tell us the updated original value of the array.
Below is the implementation of the above approach:
// C++ implementation of the approach
#include
#include
using namespace std;
// Function to count number
// of values greater or equal to x
int increaseDecreaseValue(int arr[],
int x, int n)
{
int TotalValue = 0;
for (int i = 0; i < n; i++)
{
if (arr[i] < x)
{
// Current value is less
// than required value
// then we need to increase
// the value from i + 1 to
// len(arr) by 1
for (int j = i + 1; j < n; j++)
{
arr[j] += 1;
}
}
else
{
// Current value is greater
// or equal to required
// value then we need to
// decrease the value from
// (i + 1) to len(arr)-1 by 1
TotalValue += 1;
for (int j = i + 1; j < n; j++)
{
if (arr[j] == 0)
{
continue;
}
else
{
arr[j] -= 1;
}
}
}
}
return TotalValue;
}
// Driver Code
int main()
{
int x = 4;
int arr[] = {5, 4, 3, 2, 1};
int n = sizeof(arr) / sizeof(arr[0]);
int countValue = increaseDecreaseValue(arr, x, n);
cout << countValue;
return 0;
}
Output:
1
Time Complexity: O(N)