LeetCode #528 — MEDIUM

Random Pick with Weight

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given a 0-indexed array of positive integers w where w[i] describes the weight of the i^th index.

You need to implement the function pickIndex(), which randomly picks an index in the range [0, w.length - 1] (inclusive) and returns it. The probability of picking an index i is w[i] / sum(w).

For example, if w = [1, 3], the probability of picking index 0 is 1 / (1 + 3) = 0.25 (i.e., 25%), and the probability of picking index 1 is 3 / (1 + 3) = 0.75 (i.e., 75%).

Example 1:

Input
["Solution","pickIndex"]
[[[1]],[]]
Output
[null,0]

Explanation
Solution solution = new Solution([1]);
solution.pickIndex(); // return 0. The only option is to return 0 since there is only one element in w.

Example 2:

Input
["Solution","pickIndex","pickIndex","pickIndex","pickIndex","pickIndex"]
[[[1,3]],[],[],[],[],[]]
Output
[null,1,1,1,1,0]

Explanation
Solution solution = new Solution([1, 3]);
solution.pickIndex(); // return 1. It is returning the second element (index = 1) that has a probability of 3/4.
solution.pickIndex(); // return 1
solution.pickIndex(); // return 1
solution.pickIndex(); // return 1
solution.pickIndex(); // return 0. It is returning the first element (index = 0) that has a probability of 1/4.

Since this is a randomization problem, multiple answers are allowed.
All of the following outputs can be considered correct:
[null,1,1,1,1,0]
[null,1,1,1,1,1]
[null,1,1,1,0,0]
[null,1,1,1,0,1]
[null,1,0,1,0,0]
......
and so on.

Constraints:

1 <= w.length <= 10⁴
1 <= w[i] <= 10⁵
pickIndex will be called at most 10⁴ times.

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed array of positive integers w where w[i] describes the weight of the ith index. You need to implement the function pickIndex(), which randomly picks an index in the range [0, w.length - 1] (inclusive) and returns it. The probability of picking an index i is w[i] / sum(w). For example, if w = [1, 3], the probability of picking index 0 is 1 / (1 + 3) = 0.25 (i.e., 25%), and the probability of picking index 1 is 3 / (1 + 3) = 0.75 (i.e., 75%).

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math · Binary Search

Example 1

["Solution","pickIndex"]
[[[1]],[]]

Example 2

["Solution","pickIndex","pickIndex","pickIndex","pickIndex","pickIndex"]
[[[1,3]],[],[],[],[],[]]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #528: Random Pick with Weight
class Solution {
    private int[] s;
    private Random random = new Random();

    public Solution(int[] w) {
        int n = w.length;
        s = new int[n + 1];
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + w[i];
        }
    }

    public int pickIndex() {
        int x = 1 + random.nextInt(s[s.length - 1]);
        int left = 1, right = s.length - 1;
        while (left < right) {
            int mid = (left + right) >> 1;
            if (s[mid] >= x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left - 1;
    }
}

/**
 * Your Solution object will be instantiated and called as such:
 * Solution obj = new Solution(w);
 * int param_1 = obj.pickIndex();
 */

// Accepted solution for LeetCode #528: Random Pick with Weight
type Solution struct {
	s []int
}

func Constructor(w []int) Solution {
	n := len(w)
	s := make([]int, n+1)
	for i := 0; i < n; i++ {
		s[i+1] = s[i] + w[i]
	}
	return Solution{s}
}

func (this *Solution) PickIndex() int {
	n := len(this.s)
	x := 1 + rand.Intn(this.s[n-1])
	left, right := 1, n-1
	for left < right {
		mid := (left + right) >> 1
		if this.s[mid] >= x {
			right = mid
		} else {
			left = mid + 1
		}
	}
	return left - 1
}

/**
 * Your Solution object will be instantiated and called as such:
 * obj := Constructor(w);
 * param_1 := obj.PickIndex();
 */

# Accepted solution for LeetCode #528: Random Pick with Weight
class Solution:
    def __init__(self, w: List[int]):
        self.s = [0]
        for c in w:
            self.s.append(self.s[-1] + c)

    def pickIndex(self) -> int:
        x = random.randint(1, self.s[-1])
        left, right = 1, len(self.s) - 1
        while left < right:
            mid = (left + right) >> 1
            if self.s[mid] >= x:
                right = mid
            else:
                left = mid + 1
        return left - 1


# Your Solution object will be instantiated and called as such:
# obj = Solution(w)
# param_1 = obj.pickIndex()

// Accepted solution for LeetCode #528: Random Pick with Weight
use rand::{thread_rng, Rng};

struct Solution {
    sum: Vec<i32>,
}

/**
 * `&self` means the method takes an immutable reference.
 * If you need a mutable reference, change it to `&mut self` instead.
 */
impl Solution {
    fn new(w: Vec<i32>) -> Self {
        let n = w.len();
        let mut sum = vec![0; n + 1];
        for i in 1..=n {
            sum[i] = sum[i - 1] + w[i - 1];
        }
        Self { sum }
    }

    fn pick_index(&self) -> i32 {
        let x = thread_rng().gen_range(1, self.sum.last().unwrap() + 1);
        let (mut left, mut right) = (1, self.sum.len() - 1);
        while left < right {
            let mid = (left + right) >> 1;
            if self.sum[mid] < x {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        (left - 1) as i32
    }
}

// Accepted solution for LeetCode #528: Random Pick with Weight
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #528: Random Pick with Weight
// class Solution {
//     private int[] s;
//     private Random random = new Random();
// 
//     public Solution(int[] w) {
//         int n = w.length;
//         s = new int[n + 1];
//         for (int i = 0; i < n; ++i) {
//             s[i + 1] = s[i] + w[i];
//         }
//     }
// 
//     public int pickIndex() {
//         int x = 1 + random.nextInt(s[s.length - 1]);
//         int left = 1, right = s.length - 1;
//         while (left < right) {
//             int mid = (left + right) >> 1;
//             if (s[mid] >= x) {
//                 right = mid;
//             } else {
//                 left = mid + 1;
//             }
//         }
//         return left - 1;
//     }
// }
// 
// /**
//  * Your Solution object will be instantiated and called as such:
//  * Solution obj = new Solution(w);
//  * int param_1 = obj.pickIndex();
//  */

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.