Binary Search

Introduction

When we binary search on an answer, we start with a search space of size $N$ which we know the answer lies in. Then each iteration of the binary search cuts the search space in half, so the algorithm tests $\mathcal{O}(\log N)$ values. This is efficient and much better than testing each possible value in the search space.

Binary Searching on Monotonic Functions

Let's say we have a boolean function f(x). Usually, in such problems, we want to find the maximum or minimum value of $x$ such that f(x) is true. Similarly to how binary search on an array only works on a sorted array, binary search on the answer only works if the answer function is monotonic, meaning that it is always non-decreasing or always non-increasing.

Finding The Maximum x Such That f(x) = true

We want to construct a function lastTrue such that lastTrue(lo, hi, f) returns the last x in the range [lo,hi] such that f(x) = true. If no such x exists, then lastTrue should return lo-1.

This can be done with binary search if f(x) satisfies both of the following conditions:

• If f(x) is true, then f(y) is true for all $y \leq x$.
• If f(x) is false, then f(y) is false for all $y \geq x$.

For example, if f(x) is given by the following function:

f(1) = true
f(2) = true
f(3) = true
f(4) = true
f(5) = true
f(6) = false
f(7) = false
f(8) = false

then lastTrue(1, 8, f) = 5 and lastTrue(7, 8, f) = 6.

Implementation 1

Verify that this implementation doesn't call f on any values outside of the range [lo,hi].

#include <bits/stdc++.h>
using namespace std;

int lastTrue(int lo, int hi, function<bool(int)> f) {
	for (--lo; lo < hi; ) {
		int mid = lo+(hi-lo+1)/2;
		// find the middle of the current range (rounding up)
		if (f(mid)) lo = mid;
		// if mid works, then all numbers smaller than mid also work
		else hi = mid-1;

Implementation 2

This approach is based on interval jumping. Essentially, we start from the beginning of the array, make jumps, and reduce the jump length as we get closer to the target element.

int lastTrue(int lo, int hi, function<bool(int)> f) {
	for (int dif = (hi-(--lo)); dif; dif /= 2)
		while (lo+dif <= hi && f(lo+dif)) lo += dif;
	return lo;
}

Finding The Minimum x Such That f(x) = true

We want to construct a function firstTrue such that firstTrue(lo, hi, f) returns the first x in the range [lo,hi] such that f(x) = true. If no such x exists, then firstTrue should return hi+1.

Similarly to the previous part, this can be done with binary search if f(x) satisfies both of the following conditions:

• If f(x) is true, then f(y) is true for all $y \geq x$.
• If f(x) is false, then f(y) is false for all $y \leq x$.

We will need to do the same thing, but when the condition is satisfied, we will cut the right part, and when it's not, the left part will be cut.

#include <bits/stdc++.h>
using namespace std;

int firstTrue(int lo, int hi, function<bool(int)> f) {
	for (hi ++; lo < hi; ) {
		int mid = lo+(hi-lo)/2;
		if (f(mid)) hi = mid;
		else lo = mid+1;
	}
	return lo;

Example - Maximum Median

Statement: Given an array $\texttt{arr}$ of $n$ integers, where $n$ is odd, we can perform the following operation on it $k$ times: take any element of the array and increase it by $1$. We want to make the median of the array as large as possible after $k$ operations.

Constraints: $1 \leq n \leq 2 \cdot 10^5, 1 \leq k \leq 10^9$ and $n$ is odd.

Common Mistakes

Mistake 1 - Off By One

Consider the code from CSAcademy's Binary Search on Functions.

long long f(int x) {
	return (long long)x * x;
}
int square_root(int x) {
	int left = 0, right = x;
	while (left < right) {
		int middle = (left + right) / 2;
		if (f(middle) <= x) {
			left = middle;
		} else {
			right = middle - 1;
		}
	}
	return left;
}

This results in an infinite loop if left=0 and right=1! To fix this, set middle = (left+right+1)/2 instead.

Mistake 2 - Not Accounting for Negative Bounds

Consider a slightly modified version of firstTrue:

int firstTrue(int lo, int hi, function<bool(int)> f) {
	for (hi ++; lo < hi; ) {
		int mid = (lo+hi)/2;
		if (f(mid)) hi = mid;
		else lo = mid+1;
	}
	return lo;
}

This code does not necessarily work if lo is negative! Consider the following example:

int main() {
	cout << firstTrue(-10,-10,[](int x) { return false; }) << "\n"; 
	// -8, should be -9
	cout << firstTrue(-10,-10,[](int x) { return true; }) << "\n"; 
	// infinite loop
}

This is because dividing an odd negative integer by two will round it up instead of down.

Mistake 3 - Integer Overflow

The first version of firstTrue won't work if hi-lo initially exceeds INT_MAX, while the second version of firstTrue won't work if lo+hi exceeds INT_MAX at any point during execution. If this is an issue, use long longs instead of ints.

Library Functions For Binary Search

Example - Counting Haybales

As each of the points are in the range $0 \ldots 1,000,000,000$, storing locations of haybales in a boolean array and then taking prefix sums of that would take too much time and memory.

Instead, let's place all of the locations of the haybales into a list and sort it. Now we can use binary search to count the number of cows in any range $[A,B]$ in $\mathcal{O}(\log N)$ time.

#include <bits/stdc++.h>
using namespace std;

using ll = long long;

using vi = vector<int>;
#define pb push_back
#define rsz resize
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()

#include <bits/stdc++.h>
using namespace std;

using ll = long long;

using vi = vector<int>;
#define pb push_back
#define rsz resize
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()

import java.io.*;
import java.util.*;

public class haybales{
	public static void main(String[] args) throws IOException
	{
		BufferedReader br = new BufferedReader(new FileReader(new File("haybales.in")));
		PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("haybales.out")));
		StringTokenizer st = new StringTokenizer(br.readLine());
		int N = Integer.parseInt(st.nextToken());

from bisect import bisect

inp = open("haybales.in", 'r')
out = open("haybales.out", 'w')

N, Q = map(int, inp.readline().split())
arr = sorted(list(map(int, inp.readline().split())))

for i in range(Q):
	a, b = map(int, inp.readline().split())
	print(bisect(arr, b) - bisect(arr, a-1), file=out)

inp.close()
out.close()

Problems

USACO

General