LeetCode #1093 — MEDIUM

Statistics from a Large Sample

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

The Problem

Problem Statement

You are given a large sample of integers in the range [0, 255]. Since the sample is so large, it is represented by an array count where count[k] is the number of times that k appears in the sample.

Calculate the following statistics:

minimum: The minimum element in the sample.
maximum: The maximum element in the sample.
mean: The average of the sample, calculated as the total sum of all elements divided by the total number of elements.
median:
- If the sample has an odd number of elements, then the median is the middle element once the sample is sorted.
- If the sample has an even number of elements, then the median is the average of the two middle elements once the sample is sorted.
mode: The number that appears the most in the sample. It is guaranteed to be unique.

Return the statistics of the sample as an array of floating-point numbers [minimum, maximum, mean, median, mode]. Answers within 10^-5 of the actual answer will be accepted.

Example 1:

Input: count = [0,1,3,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
Output: [1.00000,3.00000,2.37500,2.50000,3.00000]
Explanation: The sample represented by count is [1,2,2,2,3,3,3,3].
The minimum and maximum are 1 and 3 respectively.
The mean is (1+2+2+2+3+3+3+3) / 8 = 19 / 8 = 2.375.
Since the size of the sample is even, the median is the average of the two middle elements 2 and 3, which is 2.5.
The mode is 3 as it appears the most in the sample.

Example 2:

Input: count = [0,4,3,2,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
Output: [1.00000,4.00000,2.18182,2.00000,1.00000]
Explanation: The sample represented by count is [1,1,1,1,2,2,2,3,3,4,4].
The minimum and maximum are 1 and 4 respectively.
The mean is (1+1+1+1+2+2+2+3+3+4+4) / 11 = 24 / 11 = 2.18181818... (for display purposes, the output shows the rounded number 2.18182).
Since the size of the sample is odd, the median is the middle element 2.
The mode is 1 as it appears the most in the sample.

Constraints:

count.length == 256
0 <= count[i] <= 10⁹
1 <= sum(count) <= 10⁹
The mode of the sample that count represents is unique.

Step 01

Brute Force Baseline

Problem summary: You are given a large sample of integers in the range [0, 255]. Since the sample is so large, it is represented by an array count where count[k] is the number of times that k appears in the sample. Calculate the following statistics: minimum: The minimum element in the sample. maximum: The maximum element in the sample. mean: The average of the sample, calculated as the total sum of all elements divided by the total number of elements. median: If the sample has an odd number of elements, then the median is the middle element once the sample is sorted. If the sample has an even number of elements, then the median is the average of the two middle elements once the sample is sorted. mode: The number that appears the most in the sample. It is guaranteed to be unique. Return the statistics of the sample as an array of floating-point numbers [minimum, maximum, mean, median, mode]. Answers

Baseline thinking

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

Pattern signal: Array · Math

Example 1

[0,1,3,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]

Example 2

[0,4,3,2,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]

Step 02

Core Insight

What unlocks the optimal approach

The hard part is the median. Write a helper function which finds the k-th element from the sample.

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 #1093: Statistics from a Large Sample
class Solution {
    private int[] count;

    public double[] sampleStats(int[] count) {
        this.count = count;
        int mi = 1 << 30, mx = -1;
        long s = 0;
        int cnt = 0;
        int mode = 0;
        for (int k = 0; k < count.length; ++k) {
            if (count[k] > 0) {
                mi = Math.min(mi, k);
                mx = Math.max(mx, k);
                s += 1L * k * count[k];
                cnt += count[k];
                if (count[k] > count[mode]) {
                    mode = k;
                }
            }
        }
        double median
            = cnt % 2 == 1 ? find(cnt / 2 + 1) : (find(cnt / 2) + find(cnt / 2 + 1)) / 2.0;
        return new double[] {mi, mx, s * 1.0 / cnt, median, mode};
    }

    private int find(int i) {
        for (int k = 0, t = 0;; ++k) {
            t += count[k];
            if (t >= i) {
                return k;
            }
        }
    }
}

// Accepted solution for LeetCode #1093: Statistics from a Large Sample
func sampleStats(count []int) []float64 {
	find := func(i int) int {
		for k, t := 0, 0; ; k++ {
			t += count[k]
			if t >= i {
				return k
			}
		}
	}
	mi, mx := 1<<30, -1
	s, cnt, mode := 0, 0, 0
	for k, x := range count {
		if x > 0 {
			mi = min(mi, k)
			mx = max(mx, k)
			s += k * x
			cnt += x
			if x > count[mode] {
				mode = k
			}
		}
	}
	var median float64
	if cnt&1 == 1 {
		median = float64(find(cnt/2 + 1))
	} else {
		median = float64(find(cnt/2)+find(cnt/2+1)) / 2
	}
	return []float64{float64(mi), float64(mx), float64(s) / float64(cnt), median, float64(mode)}
}

# Accepted solution for LeetCode #1093: Statistics from a Large Sample
class Solution:
    def sampleStats(self, count: List[int]) -> List[float]:
        def find(i: int) -> int:
            t = 0
            for k, x in enumerate(count):
                t += x
                if t >= i:
                    return k

        mi, mx = inf, -1
        s = cnt = 0
        mode = 0
        for k, x in enumerate(count):
            if x:
                mi = min(mi, k)
                mx = max(mx, k)
                s += k * x
                cnt += x
                if x > count[mode]:
                    mode = k

        median = (
            find(cnt // 2 + 1) if cnt & 1 else (find(cnt // 2) + find(cnt // 2 + 1)) / 2
        )
        return [mi, mx, s / cnt, median, mode]

// Accepted solution for LeetCode #1093: Statistics from a Large Sample
struct Solution;

impl Solution {
    fn sample_stats(count: Vec<i32>) -> Vec<f64> {
        let mut min = std::i32::MAX;
        let mut max = std::i32::MIN;
        let total_count: i32 = count.iter().copied().sum();
        let m1 = (total_count + 1) / 2;
        let m2 = if total_count % 2 == 0 { m1 + 1 } else { m1 };
        let mut cur_count = 0;
        let mut mode_count = 0;
        let mut mode = std::i32::MAX;
        let mut sum = 0.0;
        let n = count.len();
        let mut median = 0;
        for i in 0..n {
            if count[i] > 0 {
                min = min.min(i as i32);
                max = max.max(i as i32);
                if cur_count < m1 && cur_count + count[i] >= m1 {
                    median += i;
                }
                if cur_count < m2 && cur_count + count[i] >= m2 {
                    median += i;
                }
                cur_count += count[i];
                sum += i as f64 * count[i] as f64;
                if count[i] > mode_count {
                    mode_count = count[i];
                    mode = i as i32;
                }
            }
        }
        vec![
            min as f64,
            max as f64,
            sum / total_count as f64,
            median as f64 / 2.0,
            mode as f64,
        ]
    }
}

#[test]
fn test() {
    let count = vec![
        0, 1, 3, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
    ];
    let res = vec![1.00000, 3.00000, 2.37500, 2.50000, 3.00000];
    assert_eq!(Solution::sample_stats(count), res);
}

// Accepted solution for LeetCode #1093: Statistics from a Large Sample
function sampleStats(count: number[]): number[] {
    const find = (i: number): number => {
        for (let k = 0, t = 0; ; ++k) {
            t += count[k];
            if (t >= i) {
                return k;
            }
        }
    };
    let mi = 1 << 30;
    let mx = -1;
    let [s, cnt, mode] = [0, 0, 0];
    for (let k = 0; k < count.length; ++k) {
        if (count[k] > 0) {
            mi = Math.min(mi, k);
            mx = Math.max(mx, k);
            s += k * count[k];
            cnt += count[k];
            if (count[k] > count[mode]) {
                mode = k;
            }
        }
    }
    const median =
        cnt % 2 === 1 ? find((cnt >> 1) + 1) : (find(cnt >> 1) + find((cnt >> 1) + 1)) / 2;
    return [mi, mx, s / cnt, median, mode];
}

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(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.