📖 Scenario: Imagine you are organizing a party and want to divide candies equally among children. You have a list of candy packs with different counts. You want to find the maximum number of candies each child can get so that all children get the same amount and no candy is wasted.
🎯 Goal: Build a program that uses the Binary Search on Answer Technique to find the maximum number of candies each child can get equally from the given candy packs.
📋 What You'll Learn
Create an array called candyPacks with the exact values: [10, 15, 20, 17]
Create a variable called children and set it to 5
Write a function called canDistribute that takes a number candies and returns true if it is possible to give each child candies candies from the packs, otherwise false
Use binary search between 1 and the maximum candies in any pack to find the maximum candies per child
Print the maximum candies each child can get
💡 Why This Matters
🌍 Real World
This technique helps in resource allocation problems like dividing goods, bandwidth, or tasks evenly.
💼 Career
Binary search on answer is a common pattern in coding interviews and competitive programming to optimize search over a range of answers.
Progress0 / 4 steps
1
Create the candy packs array
Create an array called candyPacks with these exact values: [10, 15, 20, 17]
DSA Typescript
Hint
Use const candyPacks: number[] = [10, 15, 20, 17]; to create the array.
2
Set the number of children
Create a variable called children and set it to 5
DSA Typescript
Hint
Use const children: number = 5; to set the number of children.
3
Write the function to check distribution possibility
Write a function called canDistribute that takes a number candies and returns true if it is possible to give each child candies candies from the packs, otherwise false. Use a for loop to sum how many children can be served with candies from each pack.
DSA Typescript
Hint
Count how many children can get candies by dividing each pack by candies and summing. Return true if count is at least children.
4
Use binary search to find maximum candies per child and print result
Use binary search between 1 and the maximum number in candyPacks to find the maximum candies each child can get. Print the result using console.log(maxCandies).
DSA Typescript
Hint
Use binary search to test mid values. If canDistribute(mid) is true, move left up and update maxCandies. Otherwise, move right down. Print maxCandies at the end.
Practice
(1/5)
1. What is the main idea behind the Binary Search on Answer technique?
easy
A. We guess an answer and check if it satisfies the problem constraints, then adjust the guess.
B. We sort the input array before applying binary search.
C. We use binary search only on sorted arrays to find elements.
D. We divide the input into two halves and solve each recursively.
Solution
Step 1: Understand the technique's purpose
Binary Search on Answer guesses a possible answer value and checks if it meets the problem's conditions.
Step 2: Adjust guess based on check
If the guess is valid, we try smaller or larger values accordingly to find the optimal answer.
Final Answer:
We guess an answer and check if it satisfies the problem constraints, then adjust the guess. -> Option A
Quick Check:
Guess and check approach = C [OK]
Hint: Think of guessing answers and verifying them stepwise [OK]
Common Mistakes:
Confusing with normal binary search on arrays
Assuming input must be sorted
Thinking it divides input recursively
2. Which of the following is the correct TypeScript syntax to start a binary search on answer between low and high?
easy
A. while (low <= high) { let mid = (low + high) / 2; }
B. for (let mid = low; mid <= high; mid++) {}
C. while (low <= high) { let mid = Math.floor((low + high) / 2); }
D. if (low < high) { let mid = (low + high) >> 1; }
Solution
Step 1: Identify correct mid calculation
Midpoint must be an integer index, so use Math.floor to avoid decimals.
Step 2: Confirm loop condition and syntax
Use while loop with condition low <= high for binary search iteration.
Final Answer:
while (low <= high) { let mid = Math.floor((low + high) / 2); } -> Option C
Quick Check:
Integer mid with while loop = B [OK]
Hint: Use Math.floor for mid to avoid decimals in binary search [OK]
Common Mistakes:
Using division without floor causing float mid
Using for loop instead of while
Using bitwise shift without understanding
3. Consider this TypeScript code snippet using binary search on answer to find minimum capacity to ship packages within D days:
function canShip(weights: number[], capacity: number, D: number): boolean {
let days = 1, total = 0;
for (const w of weights) {
if (total + w > capacity) {
days++;
total = 0;
}
total += w;
}
return days <= D;
}
function shipWithinDays(weights: number[], D: number): number {
let low = Math.max(...weights);
let high = weights.reduce((a, b) => a + b, 0);
while (low < high) {
const mid = Math.floor((low + high) / 2);
if (canShip(weights, mid, D)) high = mid;
else low = mid + 1;
}
return low;
}
console.log(shipWithinDays([1,2,3,4,5,6,7,8,9,10], 5));
What is the output?
medium
A. 15
B. 10
C. 20
D. 5
Solution
Step 1: Understand problem and code
We find minimum ship capacity to ship weights in 5 days. Minimum capacity is max weight (10), max capacity is sum (55).
Step 2: Trace binary search and canShip checks
Binary search narrows capacity to 15, which allows shipping in 5 days by grouping weights properly.
Final Answer:
15 -> Option A
Quick Check:
Minimum capacity = 15 [OK]
Hint: Check max weight as low and sum as high for binary search [OK]
Common Mistakes:
Choosing max weight instead of sum as answer
Confusing days count in canShip function
Not using Math.floor for mid calculation
4. The following TypeScript code tries to find the maximum minimum distance between elements placed in positions array using binary search on answer. Identify the bug:
function canPlace(positions: number[], distance: number, m: number): boolean {
let count = 1, lastPos = positions[0];
for (let i = 1; i < positions.length; i++) {
if (positions[i] - lastPos >= distance) {
count++;
lastPos = positions[i];
}
}
return count >= m;
}
function maxMinDistance(positions: number[], m: number): number {
positions.sort((a, b) => a - b);
let low = 0, high = positions[positions.length - 1] - positions[0];
while (low < high) {
const mid = Math.floor((low + high + 1) / 2);
if (canPlace(positions, mid, m)) low = mid;
else high = mid - 1;
}
return low;
}
medium
A. Missing increment of count inside canPlace
B. Binary search mid calculation should not add 1
C. Binary search loop condition should be low <= high
D. No bug; code works correctly
Solution
Step 1: Analyze canPlace function
It correctly counts placements with given minimum distance.
Step 2: Check binary search logic
Mid uses upper mid to avoid infinite loop, loop condition low < high is valid, and updates low/high correctly.
Final Answer:
No bug; code works correctly -> Option D
Quick Check:
Code logic correct = A [OK]
Hint: Use upper mid to avoid infinite loop in binary search [OK]
Common Mistakes:
Changing loop condition incorrectly
Removing +1 in mid causing infinite loop
Forgetting to sort positions array
5. You want to find the minimum maximum time to complete all tasks by k workers, where each worker does continuous tasks. Given tasks array of times, how would you apply binary search on answer to solve this? Choose the best approach.
hard
A. Calculate average time and assign tasks evenly without binary search.
B. Binary search on time range; check if tasks can be split into ≤ k groups with max sum ≤ mid.
C. Use binary search on number of tasks per worker directly.
D. Sort tasks and assign largest tasks first to workers.
Solution
Step 1: Define search space for answer
Minimum max time is max single task time; maximum is sum of all tasks.
Step 2: Check feasibility with mid value
Check if tasks can be divided into ≤ k continuous groups with max sum ≤ mid.
Step 3: Adjust search space based on feasibility
If feasible, try smaller max time; else increase it.
Final Answer:
Binary search on time range; check if tasks can be split into ≤ k groups with max sum ≤ mid. -> Option B
Quick Check:
Binary search on max time with group check = D [OK]
Hint: Binary search answer space, check feasibility with grouping [OK]
Common Mistakes:
Sorting tasks instead of binary searching answer
Trying to binary search on task count directly
Ignoring continuous group constraint
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