Given an array arr[] of size N, the task is to assign points to every distinct array element such that the sum of the points assigned is minimum possible.
The following conditions are required to be satisfied while assigning points:
- The minimum point that can be given to a number is 1.
- Every distinct element should be assigned a unique value.
- The points should be assigned in accordance with the value of the number, i.e. smaller numbers should be assigned lower points as compared to bigger numbers.
Examples:
Input: N = 5, arr[] = {10, 20, 10, 25}
Output: 7
Explanation: Assign 1 point to 10, 2 points to 20, 3 points to 25. Sum of points = 1 + 2 + 1 + 3 = 7, which is the minimum possible.Input: N = 4, arr[] = {7, 7, 7, 7}
Output: 4
Explanation: Assign 1 point to 7. Sum of points = 1 + 1 + 1 + 1 = 4, which is the minimum possible.
Approach: The problem can be solved greedily by assigning lower points to smaller elements. Follow the steps below to solve the problem:
- Initialize a variable ans as 1 to store the required result.
- Sort the given array in increasing order.
- Initialize a variable point as 1 to store the points of the current element.
-
Iterate in the range [0, N-1] using the variable i
- If arr[i] is equal to arr[i+1], then increment ans by point.
- Else increment the value of point by 1 and then add point to ans.
- Print the value of ans as the result.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to assign points to each // number in an array such that the // sum of points assigned is minimized void findMinimumSum( int arr[], int n)
{ // Sort the array in increasing order
sort(arr, arr + n);
// Assign point_value to 1
int point_value = 1;
// Store the required result
int total_points = 1;
// Traverse the array, arr[]
for ( int i = 0; i < n - 1; i++) {
// If adjacent elements are same add
// current point_value to total_point
if (arr[i] == arr[i + 1])
total_points += point_value;
// Otherwise, assign a new point_value
// and add it to total_points
else {
point_value = point_value + 1;
total_points += point_value;
}
}
// Print the result
cout << total_points;
} // Driver Code int main()
{ // Given Input
int arr[] = { 10, 20, 10, 25 };
int n = 4;
// Function Call
findMinimumSum(arr, n);
return 0;
} |
// Java program for the above approach import java.util.*;
class GFG{
// Function to assign points to each // number in an array such that the // sum of points assigned is minimized static void findMinimumSum( int []arr, int n)
{ // Sort the array in increasing order
Arrays.sort(arr);
// Assign point_value to 1
int point_value = 1 ;
// Store the required result
int total_points = 1 ;
// Traverse the array, arr[]
for ( int i = 0 ; i < n - 1 ; i++)
{
// If adjacent elements are same add
// current point_value to total_point
if (arr[i] == arr[i + 1 ])
total_points += point_value;
// Otherwise, assign a new point_value
// and add it to total_points
else
{
point_value = point_value + 1 ;
total_points += point_value;
}
}
// Print the result
System.out.print(total_points);
} // Driver Code public static void main(String[] args)
{ // Given Input
int []arr = { 10 , 20 , 10 , 25 };
int n = 4 ;
// Function Call
findMinimumSum(arr, n);
} } // This code is contributed by sanjoy_62. |
# Python3 program for the above approach # Function to assign points to each # number in an array such that the # sum of points assigned is minimized def findMinimumSum(arr, n):
# Sort the array in increasing order
arr = sorted (arr)
# Assign point_value to 1
point_value = 1
# Store the required result
total_points = 1
# Traverse the array, arr[]
for i in range (n - 1 ):
# If adjacent elements are same add
# current point_value to total_point
if (arr[i] = = arr[i + 1 ]):
total_points + = point_value
# Otherwise, assign a new point_value
# and add it to total_points
else :
point_value = point_value + 1
total_points + = point_value
# Print the result
print (total_points)
# Driver Code if __name__ = = '__main__' :
# Given Input
arr = [ 10 , 20 , 10 , 25 ]
n = 4
# Function Call
findMinimumSum(arr, n)
# This code is contributed by mohit kumar 29 |
// C# program for the above approach using System;
using System.Collections.Generic;
class GFG{
// Function to assign points to each // number in an array such that the // sum of points assigned is minimized static void findMinimumSum( int []arr, int n)
{ // Sort the array in increasing order
Array.Sort(arr);
// Assign point_value to 1
int point_value = 1;
// Store the required result
int total_points = 1;
// Traverse the array, arr[]
for ( int i = 0; i < n - 1; i++)
{
// If adjacent elements are same add
// current point_value to total_point
if (arr[i] == arr[i + 1])
total_points += point_value;
// Otherwise, assign a new point_value
// and add it to total_points
else
{
point_value = point_value + 1;
total_points += point_value;
}
}
// Print the result
Console.Write(total_points);
} // Driver Code public static void Main()
{ // Given Input
int []arr = { 10, 20, 10, 25 };
int n = 4;
// Function Call
findMinimumSum(arr, n);
} } // This code is contributed by SURENDRA_GANGWAR |
<script> // Javascript program for the above approach // Function to assign points to each // number in an array such that the // sum of points assigned is minimized function findMinimumSum(arr, n) {
// Sort the array in increasing order
arr.sort((a, b) => a - b);
// Assign point_value to 1
let point_value = 1;
// Store the required result
let total_points = 1;
// Traverse the array, arr[]
for (let i = 0; i < n - 1; i++) {
// If adjacent elements are same add
// current point_value to total_point
if (arr[i] == arr[i + 1])
total_points += point_value;
// Otherwise, assign a new point_value
// and add it to total_points
else {
point_value = point_value + 1;
total_points += point_value;
}
}
// Print the result
document.write(total_points);
} // Driver Code // Given Input let arr = [10, 20, 10, 25]; let n = 4; // Function Call findMinimumSum(arr, n); // This code is contributed by gfgking. </script> |
7
Time Complexity: O(N*logN)
Auxiliary Space: O(1)