Template:

Time complexity: O(log(r-l)) * O(f(m) + g(m))

Space complexity: O(1)

"""
Returns the smallest number m such that g(m) is true.
"""
def binary_search(l, r):
  while l < r:
    m = l + (r - l) // 2
    if f(m): return m    # if m is the answer
    if g(m):
      r = m              # new range [l, m)
    else
      l = m + 1          # new range [m+1, r)
  return l               # or not found

Slides:

Lower Bound / Upper Bound

#include <iostream>
#include <vector>
using namespace std;

int upper_bound(const vector<int>& A, int val, int l, int r) {
  while (l < r) {
    int m = l + (r - l) / 2;
    if (A[m] > val)
      r = m;
    else      
      l = m + 1;
  }
  return l;
}

int lower_bound(const vector<int>& A, int val, int l, int r) {
  while (l < r) {
    int m = l + (r - l) / 2;
    if (A[m] >= val)
      r = m;
    else
      l = m + 1;
  }
  return l;
}

int main() {
  vector<int> A{1, 2, 2, 2, 4, 4, 5};
  cout << lower_bound(A, 0, 0, A.size()) << endl; // 0
  cout << lower_bound(A, 2, 0, A.size()) << endl; // 1
  cout << lower_bound(A, 3, 0, A.size()) << endl; // 4
  cout << upper_bound(A, 2, 0, A.size()) << endl; // 4
  cout << upper_bound(A, 4, 0, A.size()) << endl; // 6
  cout << upper_bound(A, 5, 0, A.size()) << endl; // 7
}

Mentioned Problems

• 花花酱 LeetCode. 69 Sqrt(x)
• 花花酱 LeetCode 704. Binary Search
• 花花酱 LeetCode 875. Koko Eating Bananas
• 花花酱 LeetCode 378. Kth Smallest Element in a Sorted Matrix

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