Binary Search and Monotonic Function
In the preceeding binary search problems, the arrays are always sorted. And we stated that we can use binary search whenever you make a binary decision to shrink the search range. This might sound a little abstract. In this module, we will dive a little further into when we can use binary search.
Monotonic Function
A monotonic function is a function that is either non-decreasing or non-increasing. Given x1 and x2 where x1 > x2,
we should have f(x1) >= f(x2).
A sorted array is monotonic because the value increases or stays the same as the index increases.
If f(x) only contain boolean values True and False and think true as 1 and false as 0, then a sorted boolean array would consist of consecutive 0s and then consecutive 1s. For example, FFFFTTTTT.
Binary Search Template
feasible function
The pre-condition for binary search is to find a monotonic function f(x) that returns either True or False.
Then the problem becomes Find the First True in a Sorted Boolean Array that we already know how to solve using binary search.
We will call the function feasible to signify that whether the element at the current index is feasible (True) or not (False) to meet the problem constraints.
1def binary_search(arr: List[int], target: int) -> int:
2 left, right = 0, len(arr) - 1
3 first_true_index = -1
4 while left <= right:
5 mid = (left + right) // 2
6 if feasible(mid):
7 first_true_index = mid
8 right = mid - 1
9 else:
10 left = mid + 1
11
12 return first_true_index
131public static int binarySearch(List<Integer> arr, int target) {
2 int left = 0;
3 int right = arr.size() - 1;
4 int firstTrueIndex = -1;
5 while (left <= right) {
6 int mid = left + (right - left) / 2;
7 if (feasible(mid)) {
8 firstTrueIndex = mid;
9 right = mid - 1;
10 } else {
11 left = mid + 1;
12 }
13 }
14 return firstTrueIndex;
15}
161public static int BinarySearch(List<int> arr, int target)
2{
3 int left = 0;
4 int right = arr.Count - 1;
5 int firstTrueIndex = -1;
6 while (left <= right)
7 {
8 int mid = (left + right) / 2;
9 if (feasible(mid))
10 {
11 firstTrueIndex = mid;
12 right = mid - 1;
13 }
14 else
15 {
16 left = mid + 1;
17 }
18 }
19 return firstTrueIndex;
20}
211function binarySearch(arr, target) {
2 let left = 0;
3 let right = arr.length - 1;
4 let first_true_index = -1;
5
6 while (left <= right) {
7 let mid = Math.floor((left + right) / 2);
8 if (feasible(mid)) {
9 first_true_index = mid;
10 right = mid - 1;
11 } else {
12 left = mid + 1;
13 }
14 }
15 return first_true_index;
16}
171int binary_search(std::vector<int> arr, int target) {
2 int left = 0;
3 int right = arr.size() - 1;
4 int firstTrueIndex = -1;
5 while (left <= right) {
6 int mid = left + (right - left) / 2;
7 if (feasible(mid)) {
8 firstTrueIndex = mid;
9 right = mid - 1;
10 } else {
11 left = mid + 1;
12 }
13 }
14 return firstTrueIndex;
15}
161func binarySearch(arr []int, target int) int {
2 left := 0
3 right := len(arr) - 1
4 firstTrueIndex := -1
5 for left <= right {
6 mid := (left + right) / 2
7 if feasible(mid) {
8 firstTrueIndex = mid
9 right = mid - 1
10 } else {
11 left = mid + 1
12 }
13 }
14 return firstTrueIndex
15}
161(define (binary-search arr target)
2 (let search ([l 0]
3 [r (sub1 (vector-length arr))])
4 (if (<= l r)
5 (let* ([m (quotient (+ l r) 2)]
6 [v (vector-ref arr m)])
7 (cond
8 [(= v target) m]
9 [(< v target) (search (add1 m) r)]
10 [else (search l (sub1 r))]))
11 -1)))
121binarySearch :: Array Int Int -> Int -> Int
2binarySearch arr target = uncurry search $ bounds arr where
3 search l r
4 | l <= r = case compare (arr A.! m) target of
5 EQ -> m
6 LT -> search (succ m) r
7 GT -> search l (pred m)
8 | otherwise = -1
9 where m = l + (r - l) `div` 2
10Now the problem has become finding the feasible function and then mechanically applying the template.
In the Find the First True in a Sorted Boolean Array problem, the feasibile function is simply arr[mid] is True.
It's trickier to find the feasible function in other problems. Let's look at a few examples.
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