Java Program to Find closest number in array
Given an array of sorted integers. We need to find the closest value to the given number. Array may contain duplicate values and negative numbers.
Examples:
Input : arr[] = {1, 2, 4, 5, 6, 6, 8, 9}
Target number = 11
Output : 9
9 is closest to 11 in given array
Input :arr[] = {2, 5, 6, 7, 8, 8, 9};
Target number = 4
Output : 5
A simple solution is to traverse through the given array and keep track of absolute difference of current element with every element. Finally return the element that has minimum absolution difference.
An efficient solution is to use Binary Search.
Java
import java.util.*;
import java.lang.*;
import java.io.*;
class FindClosestNumber {
public static int findClosest( int arr[], int target)
{
int n = arr.length;
if (target <= arr[ 0 ])
return arr[ 0 ];
if (target >= arr[n - 1 ])
return arr[n - 1 ];
int i = 0 , j = n, mid = 0 ;
while (i < j) {
mid = (i + j) / 2 ;
if (arr[mid] == target)
return arr[mid];
if (target < arr[mid]) {
if (mid > 0 && target > arr[mid - 1 ])
return getClosest(arr[mid - 1 ],
arr[mid], target);
j = mid;
}
else {
if (mid < n- 1 && target < arr[mid + 1 ])
return getClosest(arr[mid],
arr[mid + 1 ], target);
i = mid + 1 ;
}
}
return arr[mid];
}
public static int getClosest( int val1, int val2,
int target)
{
if (target - val1 >= val2 - target)
return val2;
else
return val1;
}
public static void main(String[] args)
{
int arr[] = { 1 , 2 , 4 , 5 , 6 , 6 , 8 , 9 };
int target = 11 ;
System.out.println(findClosest(arr, target));
}
}
|
Output:
9
Time Complexity: O(log(n))
Auxiliary Space: O(log(n)) (implicit stack is created due to recursion)
Approach 2: Using Two Pointers
Another approach to solve this problem is to use two pointers technique, where we maintain two pointers left and right, and move them towards each other based on their absolute difference with target.
Below are the steps:
- Initialize left = 0 and right = n-1, where n is the size of the array.
- Loop while left < right
- If the absolute difference between arr[left] and target is less than or equal to the absolute difference between arr[right] and target, move left pointer one step to the right, i.e. left++
- Else, move right pointer one step to the left, i.e. right–-
- Return arr[left], which will be the element closest to the target.
Below is the implementation of the above approach:
Java
import java.util.*;
public class GFG {
public static int findClosest( int [] arr, int n,
int target)
{
int left = 0 , right = n - 1 ;
while (left < right) {
if (Math.abs(arr[left] - target)
<= Math.abs(arr[right] - target)) {
right--;
}
else {
left++;
}
}
return arr[left];
}
public static void main(String[] args)
{
int [] arr = { 1 , 2 , 4 , 5 , 6 , 6 , 8 , 8 , 9 };
int n = arr.length;
int target = 11 ;
System.out.println(findClosest(arr, n, target));
}
}
|
Time Complexity: O(N), where n is the length of the array.
Auxiliary Space: O(1)
Please refer complete article on Find closest number in array for more details!
Last Updated :
13 Apr, 2023
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