Binary Search on the Answer

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 $O(\log N)$ values. This is efficient and much better than testing each possible value in the search space.

Let's say we have a function check(x) that returns true if the answer of $x$ is possible, and false otherwise. Usually, in such problems, we want to find the maximum or minimum value of $x$ such that check(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.

In particular, if we want to find the maximum x such that check(x) is true, then we can binary search if check(x) satisfies both of the following conditions:

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

For example, the last point at which the condition below is satisfied is 5.

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

Below, we present several algorithms for binary search, which search for the maximum x such that f(x) is true.

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

int lstTrue(function<bool(int)> f, int lo, int hi) {
    int res = lo-1;
    while (lo <= hi) {
        int mid = (lo+hi)/2; // find the middle of the current range
        if (f(mid)) {
            // if mid works, then all numbers smaller than mid also work
            // so we only care about the part after mid

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

public class test{
   public static boolean f(int x){
      return x*x <= 30;
      //function f(x) returns true or false
   }
   public static int lstTrue(int lo, int hi) {
      int res = lo-1;

We can shorten the function by removing res and using a ternary if expression.

int lstTrue(function<bool(int)> f, int lo, int hi) {
    for (lo --; lo < hi; ) {
        int mid = (lo+hi+1)/2;
        f(mid) ? lo = mid : hi = mid-1;
    }
    return lo;
}

public static int lstTrue(int lo, int hi) {
   for (lo --; lo < hi; ) {
      int mid = (lo+hi+1)/2;
      if(f(mid)) lo = mid; else hi = mid-1;
   }
   return lo;
}

There is also another approach to binary searching 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.

long long search(){
    long long pos = 0; long long max = 2E9;
    for(long long a = max; a >= 1; a /= 2){
        while(check(pos+a)) pos += a;
    }
    return pos;
}

static long search(){
    long pos = 0; long max = (long)2E9;
    for(long a = max; a >= 1; a /= 2){
        while(check(pos+a)) pos += a;
    }
    return pos;
}

If instead we're looking for the minimum x that satisfies some condition, then we can binary search if check(x) satisfies both of the following conditions:

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

The binary search function for this is very similar. 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 fstTrue(function<bool(int)> f, int lo, int hi) {
    // returns smallest x in [lo,hi] that satisfies f
    // hi+1 if no x satisfies f
    for (hi ++; lo < hi; ) {
        int mid = (lo+hi)/2;
        f(mid) ? hi = mid : lo = mid+1;
    }

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

public class test{
   public static boolean f(int x){
      return x*x >= 30;
      //function f(x) returns true or false
   }
   public static int fstTrue(int lo, int hi) {
      for (hi ++; lo < hi; ) {

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

public static void main(String[] args) throws IOException
{
     System.out.println(fstTrue(-10,-7)); // infinite loop
}

int fstTrue(function<bool(int)> f, int lo, int hi) {
    for (hi ++; lo < hi; ) {
        int mid = lo+hi; mid = (mid-(mid&1))/2;
        f(mid) ? hi = mid : lo = mid+1;
    }
    return lo;
}

public static int fstTrue(int lo, int hi) {
   for (hi ++; lo < hi; ) {
      // returns smallest x in [lo,hi] that satisfies f
      // hi+1 if no x satisfies f
      int mid = lo+hi; 
      mid = (mid-(mid&1))/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.

Problems

USACO

General