C++ Program to Find closest number in array
Last Updated :
03 Apr, 2023
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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.
// CPP program to find element
// closest to given target.
#include <bits/stdc++.h>
using namespace std;
int getClosest(int, int, int);
// Returns element closest to target in arr[]
int findClosest(int arr[], int n, int target)
{
// Corner cases
if (target <= arr[0])
return arr[0];
if (target >= arr[n - 1])
return arr[n - 1];
// Doing binary search
int i = 0, j = n, mid = 0;
while (i < j) {
mid = (i + j) / 2;
if (arr[mid] == target)
return arr[mid];
/* If target is less than array element,
then search in left */
if (target < arr[mid]) {
// If target is greater than previous
// to mid, return closest of two
if (mid > 0 && target > arr[mid - 1])
return getClosest(arr[mid - 1],
arr[mid], target);
/* Repeat for left half */
j = mid;
}
// If target is greater than mid
else {
if (mid < n - 1 && target < arr[mid + 1])
return getClosest(arr[mid],
arr[mid + 1], target);
// update i
i = mid + 1;
}
}
// Only single element left after search
return arr[mid];
}
// Method to compare which one is the more close.
// We find the closest by taking the difference
// between the target and both values. It assumes
// that val2 is greater than val1 and target lies
// between these two.
int getClosest(int val1, int val2,
int target)
{
if (target - val1 >= val2 - target)
return val2;
else
return val1;
}
// Driver code
int main()
{
int arr[] = { 1, 2, 4, 5, 6, 6, 8, 9 };
int n = sizeof(arr) / sizeof(arr[0]);
int target = 11;
cout << (findClosest(arr, n, target));
}
// This code is contributed by Smitha Dinesh Semwal
// CPP program to find element// closest to given target.using namespace std;int getClosest(int, int, int);// Returns element closest to target in arr[]int findClosest(int arr[], int n, int target){ // Corner cases if (target <= arr[0]) return arr[0]; if (target >= arr[n - 1]) return arr[n - 1]; // Doing binary search int i = 0, j = n, mid = 0; while (i < j) { mid = (i + j) / 2; if (arr[mid] == target) return arr[mid]; /* If target is less than array element, then search in left */ if (target < arr[mid]) { // If target is greater than previous // to mid, return closest of two if (mid > 0 && target > arr[mid - 1]) return getClosest(arr[mid - 1], arr[mid], target); /* Repeat for left half */ j = mid; } // If target is greater than mid else { if (mid < n - 1 && target < arr[mid + 1]) return getClosest(arr[mid], arr[mid + 1], target); // update i i = mid + 1; } } // Only single element left after search return arr[mid];}// Method to compare which one is the more close.// We find the closest by taking the difference// between the target and both values. It assumes// that val2 is greater than val1 and target lies// between these two.int getClosest(int val1, int val2, int target){ if (target - val1 >= val2 - target) return val2; else return val1;}// Driver codeint main(){ int arr[] = { 1, 2, 4, 5, 6, 6, 8, 9 }; int n = sizeof(arr) / sizeof(arr[0]); int target = 11; cout << (findClosest(arr, n, target));}// This code is contributed by Smitha Dinesh SemwalOutput:
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:
// CPP program to find element
// closest to given target using two pointers
#include <bits/stdc++.h>
using namespace std;
int findClosest(int arr[], int n, int target)
{
int left = 0, right = n - 1;
while (left < right) {
if (abs(arr[left] - target)
<= abs(arr[right] - target)) {
right--;
}
else {
left++;
}
}
return arr[left];
}
int main()
{
int arr[] = { 1, 2, 4, 5, 6, 6, 8, 8, 9 };
int n = sizeof(arr) / sizeof(arr[0]);
int target = 11;
cout << findClosest(arr, n, target);
return 0;
}
// This code is contributed by Susobhan Akhuli
// CPP program to find element// closest to given target using two pointersusing namespace std;int findClosest(int arr[], int n, int target){ int left = 0, right = n - 1; while (left < right) { if (abs(arr[left] - target) <= abs(arr[right] - target)) { right--; } else { left++; } } return arr[left];}int main(){ int arr[] = { 1, 2, 4, 5, 6, 6, 8, 8, 9 }; int n = sizeof(arr) / sizeof(arr[0]); int target = 11; cout << findClosest(arr, n, target); return 0;}// This code is contributed by Susobhan AkhuliOutput
9
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!
