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.
Method #1: Using Binary search
Python3
# Python3 program to find element# closest to given target.# Returns element closest to target in arr[]def findClosest(arr, n, target): # Corner cases if (target <= arr[0]): return arr[0] if (target >= arr[n - 1]): return arr[n - 1] # Doing binary search 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 and 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 and 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.def getClosest(val1, val2, target): if (target - val1 >= val2 - target): return val2 else: return val1# Driver codearr = [1, 2, 4, 5, 6, 6, 8, 9]n = len(arr)target = 11print(findClosest(arr, n, target)) |
9
Time Complexity: O(log(n))
Auxiliary Space: O(log(n)) (implicit stack is created due to recursion)
Method #2: Using min() function
Python3
# Python3 program to find element# closest to given target.def findClosestValue(givenList, target): def difference(givenList): return abs(givenList - target) result = min(givenList, key=difference) return resultif __name__ == "__main__": givenList = [1, 2, 4, 5, 6, 6, 8, 9] target = 11 result = findClosestValue(givenList, target) print("The closest value to the " + str(target)+" is", result)# This code is contributed by vikkycirus |
The closest value to the 11 is 9
Method #3: 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:
Python3
# Python program to find element# closest to given target using two pointersimport sys def findClosest(arr, n, target): left, right = 0, n - 1 while left < right: if abs(arr[left] - target) <= abs(arr[right] - target): right -= 1 else: left += 1 return arr[left] if __name__ == "__main__": arr = [1, 2, 4, 5, 6, 6, 8, 8, 9] n = len(arr) target = 11 print(findClosest(arr, n, target))# This code is contributed by Susobhan Akhuli |
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!
