# Closest value to K from an unsorted array

• Difficulty Level : Easy
• Last Updated : 05 May, 2021

Given an array arr[] consisting of N integers and an integer K, the task is to find the array element closest to K. If multiple closest values exist, then print the smallest one.

Examples:

Input: arr[]={4, 2, 8, 11, 7}, K = 6
Output: 7
Explanation:
The absolute difference between 4 and 6 is |4 – 6| = 2
The absolute difference between 2 and 6 is |2 – 6| = 4
The absolute difference between 8 and 6 is |8– 6| = 2
The absolute difference between 11 and 6 is |11 – 6| = 5
The absolute difference between 7 and 6 is |7 – 6| = 1
Here, the absolute difference between K(=6) and 7 is minimum. Therefore, the closest value of K(=6) is 7

Input: arr[]={100, 200, 400}, K = 300
Output: 200

Approach: The idea is to traverse the given array and print an element of the array which gives the minimum absolute difference with the given integer K. Follow the steps below to solve the problem:

1. Initialize a variable, say res, to store the array element the closest value to K.
2. Traverse the array and compare the absolute value of abs(K – res) and the absolute value of abs(K – arr[i]).
3. If the value of abs(K – res) exceeds abs(K – arr[i]), then update res to arr[i].
4. Finally, print res.

Below is the implementation of the above approach:

## C++

 `// C++ program to implement``// the above approach` `#include ``using` `namespace` `std;` `// Function to get``// the closest value``int` `clostVal(``int` `arr[],``             ``int` `N,``             ``int` `K)``{``    ``// Stores the closest``    ``// value to K``    ``int` `res = arr;` `    ``// Traverse the array``    ``for` `(``int` `i = 1; i < N;``         ``i++) {` `        ``// If absolute difference``        ``// of K and res exceeds``        ``// absolute difference of K``        ``// and current element``        ``if` `(``abs``(K - res) > ``abs``(K - arr[i])) {``            ``res = arr[i];``        ``}``    ``}` `    ``// Return the closest``    ``// array element``    ``return` `res;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 100, 200, 400 };``    ``int` `K = 300;``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` `    ``cout << clostVal(arr, N, K);``}`

## Java

 `// Java program to implement``// the above approach``import` `java.io.*;` `class` `GFG{`` ` `// Function to get``// the closest value``static` `int` `clostVal(``int` `arr[], ``int` `N,``                    ``int` `K)``{``    ` `    ``// Stores the closest``    ``// value to K``    ``int` `res = arr[``0``];`` ` `    ``// Traverse the array``    ``for``(``int` `i = ``1``; i < N; i++)``    ``{``        ` `        ``// If absolute difference``        ``// of K and res exceeds``        ``// absolute difference of K``        ``// and current element``        ``if` `(Math.abs(K - res) >``            ``Math.abs(K - arr[i]))``        ``{``            ``res = arr[i];``        ``}``    ``}`` ` `    ``// Return the closest``    ``// array element``    ``return` `res;``}`` ` `// Driver Code``public` `static` `void` `main (String[] args)``{``    ``int` `arr[] = { ``100``, ``200``, ``400` `};``    ``int` `K = ``300``;``    ``int` `N = arr.length;``    ` `    ``System.out.print(clostVal(arr, N, K));``}``}` `// This code is contributed by code_hunt`

## Python3

 `# Python3 program to implement``# the above approach` `# Function to get``# the closest value``def` `clostVal(arr, N, K):``    ` `    ``# Stores the closest``    ``# value to K``    ``res ``=` `arr[``0``]` `    ``# Traverse the array``    ``for` `i ``in` `range``(``1``, N, ``1``):` `        ``# If absolute difference``        ``# of K and res exceeds``        ``# absolute difference of K``        ``# and current element``        ``if` `(``abs``(K ``-` `res) >``            ``abs``(K ``-` `arr[i])):``            ``res ``=` `arr[i]` `    ``# Return the closest``    ``# array element``    ``return` `res` `# Driver Code``arr ``=` `[ ``100``, ``200``, ``400` `]``K ``=` `300``N ``=` `len``(arr)` `print``(clostVal(arr, N, K))` `# This code is contributed by susmitakundugoaldanga`

## C#

 `// C# program to implement``// the above approach ``using` `System;` `class` `GFG{`` ` `// Function to get``// the closest value``static` `int` `clostVal(``int``[] arr, ``int` `N,``                    ``int` `K)``{``    ` `    ``// Stores the closest``    ``// value to K``    ``int` `res = arr;`` ` `    ``// Traverse the array``    ``for``(``int` `i = 1; i < N; i++)``    ``{``        ` `        ``// If absolute difference``        ``// of K and res exceeds``        ``// absolute difference of K``        ``// and current element``        ``if` `(Math.Abs(K - res) >``            ``Math.Abs(K - arr[i]))``        ``{``            ``res = arr[i];``        ``}``    ``}`` ` `    ``// Return the closest``    ``// array element``    ``return` `res;``}`` ` `// Driver Code``public` `static` `void` `Main ()``{``    ``int``[] arr = { 100, 200, 400 };``    ``int` `K = 300;``    ``int` `N = arr.Length;``    ` `    ``Console.WriteLine(clostVal(arr, N, K));``}``}` `// This code is contributed by sanjoy_62`

## Javascript

 ``

Output:

`200`

Time Complexity: O(N)
Auxiliary Space: O(1)

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