Given a function F(n) = P – (0.006 * n), where P is given. Given a list of integers and a number, 
. The task is to find the number from the given list for which the value of the function is closest to .
Examples:
Input : P = 12, A = 5
List = {1000, 2000}
Output : 1
Explanation :
Given, P=12, A=5
For 1000, F(1000) is 12 - 1000×0.006 = 6
For 2000, F(2000) is 12 - 2000×0.006 = 0
As the nearest value to 5 is 6,
so the answer is 1000.
Input : P = 21, A = -11
List = {81234, 94124, 52141}
Output : 3
Approach: Iterate over each value in the given list and find F(n) for every value. Now, compare the absolute difference of every value of F(n) and A and the value of 
, for which the absolute difference is minimum is the answer.
Below is the implementation of the above approach:
C++
// C++ program to find number from// given list for which value of the// function is closest to A#include <bits/stdc++.h>using namespace std;// Function to find number from// given list for which value of the// function is closest to Aint leastValue(int P, int A, int N, int a[]){ // Stores the final index int ans = -1; // Declaring a variable to store // the minimum absolute difference float tmp = (float)INFINITY; for (int i = 0; i < N; i++) { // Finding F(n) float t = P - a[i] * 0.006; // Updating the index of the answer if // new absolute difference is less than tmp if (abs(t-A) < tmp) { tmp = abs(t - A); ans = i; } } return a[ans];}// Driver codeint main(){ int N = 2, P = 12, A = 2005; int a[] = {1000, 2000}; cout << leastValue(P, A, N, a) << endl;}// This code is contributed by// sanjeev2552 |
Java
// Java program to find number from// given list for which value of the// function is closest to Aimport java.util.*;class GFG{// Function to find number from// given list for which value of the// function is closest to Astatic int leastValue(int P, int A, int N, int a[]){ // Stores the final index int ans = -1; // Declaring a variable to store // the minimum absolute difference float tmp = Float.MAX_VALUE; for (int i = 0; i < N; i++) { // Finding F(n) float t = (float) (P - a[i] * 0.006); // Updating the index of the answer if // new absolute difference is less than tmp if (Math.abs(t-A) < tmp) { tmp = Math.abs(t - A); ans = i; } } return a[ans];}// Driver codepublic static void main(String[] args){ int N = 2, P = 12, A = 2005; int a[] = {1000, 2000}; System.out.println(leastValue(P, A, N, a));}}// This code is contributed by 29AjayKumar |
Python3
# Python program to find number from # given list for which value of the # function is closest to A# Function to find number from # given list for which value of the # function is closest to Adef leastValue(P, A, N, a): # Stores the final index ans = -1 # Declaring a variable to store # the minimum absolute difference tmp = float('inf') for i in range(N): # Finding F(n) t = P - a[i] * 0.006 # Updating the index of the answer if # new absolute difference is less than tmp if abs(t - A) < tmp: tmp = abs(t - A) ans = i return a[ans]# Driver CodeN, P, A = 2, 12, 5a = [1000, 2000]print(leastValue(P, A, N, a)) |
C#
// C# program to find number from// given list for which value of the// function is closest to Ausing System; class GFG{// Function to find number from// given list for which value of the// function is closest to Astatic int leastValue(int P, int A, int N, int []a){ // Stores the final index int ans = -1; // Declaring a variable to store // the minimum absolute difference float tmp = float.MaxValue; for (int i = 0; i < N; i++) { // Finding F(n) float t = (float) (P - a[i] * 0.006); // Updating the index of the answer if // new absolute difference is less than tmp if (Math.Abs(t-A) < tmp) { tmp = Math.Abs(t - A); ans = i; } } return a[ans];}// Driver codepublic static void Main(String[] args){ int N = 2, P = 12, A = 2005; int []a = {1000, 2000}; Console.WriteLine(leastValue(P, A, N, a));}}// This code is contributed by Rajput-Ji |
PHP
<?php// PHP program to find number from // given list for which value of the // function is closest to A// Function to find number from // given list for which value of the // function is closest to Afunction leastValue($P, $A, $N, $a){ // Stores the final index $ans = -1; // Declaring a variable to store // the minimum absolute difference $tmp = PHP_INT_MAX; for($i = 0; $i < $N; $i++) { // Finding F(n) $t = $P - $a[$i] * 0.006; // Updating the index of the // answer if new absolute // difference is less than tmp if (abs($t - $A) < $tmp) { $tmp = abs($t - $A); $ans = $i; } } return $a[$ans];}// Driver Code$N = 2;$P = 12;$A = 5;$a = array(1000, 2000);print(leastValue($P, $A, $N, $a));// This code is contributed by mits?> |
Javascript
<script>// Javascript program to find number from// given list for which value of the// function is closest to A // Function to find number from// given list for which value of the// function is closest to Afunction leastValue(P, A, N, a){ // Stores the final index let ans = -1; // Declaring a variable to store // the minimum absolute difference let tmp = Number.MAX_VALUE; for (let i = 0; i < N; i++) { // Finding F(n) let t = (P - a[i] * 0.006); // Updating the index of the answer if // new absolute difference is less than tmp if (Math.abs(t-A) < tmp) { tmp = Math.abs(t - A); ans = i; } } return a[ans];} // Driver code let N = 2, P = 12, A = 2005; let a = [1000, 2000]; document.write(leastValue(P, A, N, a)) </script> |
Output:
1000
Time Complexity: O(N), where N represents the size of the array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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