LeetCode #398 — MEDIUM

Random Pick Index

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

The Problem

Problem Statement

Given an integer array nums with possible duplicates, randomly output the index of a given target number. You can assume that the given target number must exist in the array.

Implement the Solution class:

Solution(int[] nums) Initializes the object with the array nums.
int pick(int target) Picks a random index i from nums where nums[i] == target. If there are multiple valid i's, then each index should have an equal probability of returning.

Example 1:

Input
["Solution", "pick", "pick", "pick"]
[[[1, 2, 3, 3, 3]], [3], [1], [3]]
Output
[null, 4, 0, 2]

Explanation
Solution solution = new Solution([1, 2, 3, 3, 3]);
solution.pick(3); // It should return either index 2, 3, or 4 randomly. Each index should have equal probability of returning.
solution.pick(1); // It should return 0. Since in the array only nums[0] is equal to 1.
solution.pick(3); // It should return either index 2, 3, or 4 randomly. Each index should have equal probability of returning.

Constraints:

1 <= nums.length <= 2 * 10⁴
-2³¹ <= nums[i] <= 2³¹ - 1
target is an integer from nums.
At most 10⁴ calls will be made to pick.

Step 01

Brute Force Baseline

Problem summary: Given an integer array nums with possible duplicates, randomly output the index of a given target number. You can assume that the given target number must exist in the array. Implement the Solution class: Solution(int[] nums) Initializes the object with the array nums. int pick(int target) Picks a random index i from nums where nums[i] == target. If there are multiple valid i's, then each index should have an equal probability of returning.

Baseline thinking

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

Pattern signal: Hash Map · Math

Example 1

["Solution","pick","pick","pick"]
[[[1,2,3,3,3]],[3],[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 #398: Random Pick Index
class Solution {
    private int[] nums;
    private Random random = new Random();

    public Solution(int[] nums) {
        this.nums = nums;
    }

    public int pick(int target) {
        int n = 0, ans = 0;
        for (int i = 0; i < nums.length; ++i) {
            if (nums[i] == target) {
                ++n;
                int x = 1 + random.nextInt(n);
                if (x == n) {
                    ans = i;
                }
            }
        }
        return ans;
    }
}

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

// Accepted solution for LeetCode #398: Random Pick Index
type Solution struct {
	nums []int
}

func Constructor(nums []int) Solution {
	return Solution{nums}
}

func (this *Solution) Pick(target int) int {
	n, ans := 0, 0
	for i, v := range this.nums {
		if v == target {
			n++
			x := 1 + rand.Intn(n)
			if n == x {
				ans = i
			}
		}
	}
	return ans
}

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

# Accepted solution for LeetCode #398: Random Pick Index
class Solution:
    def __init__(self, nums: List[int]):
        self.nums = nums

    def pick(self, target: int) -> int:
        n = ans = 0
        for i, v in enumerate(self.nums):
            if v == target:
                n += 1
                x = random.randint(1, n)
                if x == n:
                    ans = i
        return ans


# Your Solution object will be instantiated and called as such:
# obj = Solution(nums)
# param_1 = obj.pick(target)

// Accepted solution for LeetCode #398: Random Pick Index
use rand::prelude::*;

struct Solution {
    rng: ThreadRng,
    nums: Vec<i32>,
}

impl Solution {
    fn new(nums: Vec<i32>) -> Self {
        let rng = thread_rng();
        Solution { rng, nums }
    }
    fn pick(&mut self, target: i32) -> i32 {
        let n = self.nums.len();
        let mut count = 0;
        let mut res = 0;
        for i in 0..n {
            if self.nums[i] == target {
                count += 1;
                if self.rng.gen_range(0, count) == 0 {
                    res = i;
                }
            }
        }
        res as i32
    }
}

// Accepted solution for LeetCode #398: Random Pick Index
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #398: Random Pick Index
// class Solution {
//     private int[] nums;
//     private Random random = new Random();
// 
//     public Solution(int[] nums) {
//         this.nums = nums;
//     }
// 
//     public int pick(int target) {
//         int n = 0, ans = 0;
//         for (int i = 0; i < nums.length; ++i) {
//             if (nums[i] == target) {
//                 ++n;
//                 int x = 1 + random.nextInt(n);
//                 if (x == n) {
//                     ans = i;
//                 }
//             }
//         }
//         return ans;
//     }
// }
// 
// /**
//  * Your Solution object will be instantiated and called as such:
//  * Solution obj = new Solution(nums);
//  * int param_1 = obj.pick(target);
//  */

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(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.

HASH MAP

O(n) time

O(n) space

One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.

Shortcut: Need to check “have I seen X before?” → hash map → O(n) time, O(n) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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.