LeetCode #3821 — HARD

Find Nth Smallest Integer With K One Bits

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given two positive integers n and k.

Return an integer denoting the n^th smallest positive integer that has exactly k ones in its binary representation. It is guaranteed that the answer is strictly less than 2⁵⁰.

Example 1:

Input: n = 4, k = 2

Output: 9

Explanation:

The 4 smallest positive integers that have exactly k = 2 ones in their binary representations are:

3 = 11₂
5 = 101₂
6 = 110₂
9 = 1001₂

Example 2:

Input: n = 3, k = 1

Output: 4

Explanation:

The 3 smallest positive integers that have exactly k = 1 one in their binary representations are:

1 = 1₂
2 = 10₂
4 = 100₂

Constraints:

1 <= n <= 2⁵⁰
1 <= k <= 50
The answer is strictly less than 2⁵⁰.

Step 01

Brute Force Baseline

Problem summary: You are given two positive integers n and k. Return an integer denoting the nth smallest positive integer that has exactly k ones in its binary representation. It is guaranteed that the answer is strictly less than 250.

Baseline thinking

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

Pattern signal: Math · Bit Manipulation

Example 1

4
2

Example 2

3
1

Step 02

Core Insight

What unlocks the optimal approach

Since the answer is strictly less than <code>2<sup>50</sup></code>, we can iterate over the number of bits (the length) of the result.
Precompute binomial coefficients <code>C(n, k)</code> to count how many numbers with exactly <code>k</code> set bits exist for a given length.
Determine the position of the most significant bit first, then greedily determine the remaining bits from MSB to LSB based on the remaining count <code>n</code>.

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

Largest constraint values

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 #3821: Find Nth Smallest Integer With K One Bits
class Solution {
    private static final int MX = 50;
    private static final long[][] c = new long[MX][MX + 1];

    static {
        for (int i = 0; i < MX; i++) {
            c[i][0] = 1;
            for (int j = 1; j <= i; j++) {
                c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
            }
        }
    }

    public long nthSmallest(long n, int k) {
        long ans = 0;
        for (int i = 49; i >= 0; i--) {
            if (n > c[i][k]) {
                n -= c[i][k];
                ans |= 1L << i;
                k--;
                if (k == 0) {
                    break;
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3821: Find Nth Smallest Integer With K One Bits
const MX = 50

var c [MX][MX + 1]int64

func init() {
	for i := 0; i < MX; i++ {
		c[i][0] = 1
		for j := 1; j <= i; j++ {
			c[i][j] = c[i-1][j-1] + c[i-1][j]
		}
	}
}

func nthSmallest(n int64, k int) int64 {
	var ans int64 = 0
	for i := 49; i >= 0; i-- {
		if n > c[i][k] {
			n -= c[i][k]
			ans |= 1 << i
			k--
			if k == 0 {
				break
			}
		}
	}
	return ans
}

# Accepted solution for LeetCode #3821: Find Nth Smallest Integer With K One Bits
mx = 50
c = [[0] * (mx + 1) for _ in range(mx)]
for i in range(mx):
    c[i][0] = 1
    for j in range(1, i + 1):
        c[i][j] = c[i - 1][j - 1] + c[i - 1][j]


class Solution:
    def nthSmallest(self, n: int, k: int) -> int:
        ans = 0
        for i in range(49, -1, -1):
            if n > c[i][k]:
                n -= c[i][k]
                ans |= 1 << i
                k -= 1
                if k == 0:
                    break
        return ans

// Accepted solution for LeetCode #3821: Find Nth Smallest Integer With K One Bits
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3821: Find Nth Smallest Integer With K One Bits
// class Solution {
//     private static final int MX = 50;
//     private static final long[][] c = new long[MX][MX + 1];
// 
//     static {
//         for (int i = 0; i < MX; i++) {
//             c[i][0] = 1;
//             for (int j = 1; j <= i; j++) {
//                 c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
//             }
//         }
//     }
// 
//     public long nthSmallest(long n, int k) {
//         long ans = 0;
//         for (int i = 49; i >= 0; i--) {
//             if (n > c[i][k]) {
//                 n -= c[i][k];
//                 ans |= 1L << i;
//                 k--;
//                 if (k == 0) {
//                     break;
//                 }
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3821: Find Nth Smallest Integer With K One Bits
const MX = 50;
const c: bigint[][] = Array.from({ length: MX }, () => Array(MX + 1).fill(0n));

for (let i = 0; i < MX; i++) {
    c[i][0] = 1n;
    for (let j = 1; j <= i; j++) {
        c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
    }
}

function nthSmallest(n: number, k: number): number {
    let nn = BigInt(n);
    let ans = 0n;
    for (let i = 49; i >= 0; i--) {
        if (nn > c[i][k]) {
            nn -= c[i][k];
            ans |= 1n << BigInt(i);
            if (--k === 0) {
                break;
            }
        }
    }
    return Number(ans);
}

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^2 M)

Space

O(log^2 M)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.