LeetCode #3116 — HARD

Kth Smallest Amount With Single Denomination Combination

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

The Problem

Problem Statement

You are given an integer array coins representing coins of different denominations and an integer k.

You have an infinite number of coins of each denomination. However, you are not allowed to combine coins of different denominations.

Return the k^th smallest amount that can be made using these coins.

Example 1:

Input: coins = [3,6,9], k = 3

Output: 9

Explanation: The given coins can make the following amounts:
Coin 3 produces multiples of 3: 3, 6, 9, 12, 15, etc.
Coin 6 produces multiples of 6: 6, 12, 18, 24, etc.
Coin 9 produces multiples of 9: 9, 18, 27, 36, etc.
All of the coins combined produce: 3, 6, 9, 12, 15, etc.

Example 2:

Input: coins = [5,2], k = 7

Output: 12

Explanation: The given coins can make the following amounts:
Coin 5 produces multiples of 5: 5, 10, 15, 20, etc.
Coin 2 produces multiples of 2: 2, 4, 6, 8, 10, 12, etc.
All of the coins combined produce: 2, 4, 5, 6, 8, 10, 12, 14, 15, etc.

Constraints:

1 <= coins.length <= 15
1 <= coins[i] <= 25
1 <= k <= 2 * 10⁹
coins contains pairwise distinct integers.

Step 01

Brute Force Baseline

Problem summary: You are given an integer array coins representing coins of different denominations and an integer k. You have an infinite number of coins of each denomination. However, you are not allowed to combine coins of different denominations. Return the kth smallest amount that can be made using these coins.

Baseline thinking

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

Pattern signal: Array · Math · Binary Search · Bit Manipulation

Example 1

[3,6,9]
3

Example 2

[5,2]
7

Core Insight

What unlocks the optimal approach

Binary search the answer <code>x</code>.
Use the inclusion-exclusion principle to count the number of distinct amounts that can be made up to <code>x</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 #3116: Kth Smallest Amount With Single Denomination Combination
class Solution {
    private int[] coins;
    private int k;

    public long findKthSmallest(int[] coins, int k) {
        this.coins = coins;
        this.k = k;
        long l = 1, r = (long) 1e11;
        while (l < r) {
            long mid = (l + r) >> 1;
            if (check(mid)) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }

    private boolean check(long mx) {
        long cnt = 0;
        int n = coins.length;
        for (int i = 1; i < 1 << n; ++i) {
            long v = 1;
            for (int j = 0; j < n; ++j) {
                if ((i >> j & 1) == 1) {
                    v = lcm(v, coins[j]);
                    if (v > mx) {
                        break;
                    }
                }
            }
            int m = Integer.bitCount(i);
            if (m % 2 == 1) {
                cnt += mx / v;
            } else {
                cnt -= mx / v;
            }
        }
        return cnt >= k;
    }

    private long lcm(long a, long b) {
        return a * b / gcd(a, b);
    }

    private long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a % b);
    }
}

// Accepted solution for LeetCode #3116: Kth Smallest Amount With Single Denomination Combination
func findKthSmallest(coins []int, k int) int64 {
	var r int = 1e11
	n := len(coins)
	ans := sort.Search(r, func(mx int) bool {
		cnt := 0
		for i := 1; i < 1<<n; i++ {
			v := 1
			for j, x := range coins {
				if i>>j&1 == 1 {
					v = lcm(v, x)
					if v > mx {
						break
					}
				}
			}
			m := bits.OnesCount(uint(i))
			if m%2 == 1 {
				cnt += mx / v
			} else {
				cnt -= mx / v
			}
		}
		return cnt >= k
	})
	return int64(ans)
}

func gcd(a, b int) int {
	if b == 0 {
		return a
	}
	return gcd(b, a%b)
}

func lcm(a, b int) int {
	return a * b / gcd(a, b)
}

# Accepted solution for LeetCode #3116: Kth Smallest Amount With Single Denomination Combination
class Solution:
    def findKthSmallest(self, coins: List[int], k: int) -> int:
        def check(mx: int) -> bool:
            cnt = 0
            for i in range(1, 1 << len(coins)):
                v = 1
                for j, x in enumerate(coins):
                    if i >> j & 1:
                        v = lcm(v, x)
                        if v > mx:
                            break
                m = i.bit_count()
                if m & 1:
                    cnt += mx // v
                else:
                    cnt -= mx // v
            return cnt >= k

        return bisect_left(range(10**11), True, key=check)

// Accepted solution for LeetCode #3116: Kth Smallest Amount With Single Denomination Combination
// 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 #3116: Kth Smallest Amount With Single Denomination Combination
// class Solution {
//     private int[] coins;
//     private int k;
// 
//     public long findKthSmallest(int[] coins, int k) {
//         this.coins = coins;
//         this.k = k;
//         long l = 1, r = (long) 1e11;
//         while (l < r) {
//             long mid = (l + r) >> 1;
//             if (check(mid)) {
//                 r = mid;
//             } else {
//                 l = mid + 1;
//             }
//         }
//         return l;
//     }
// 
//     private boolean check(long mx) {
//         long cnt = 0;
//         int n = coins.length;
//         for (int i = 1; i < 1 << n; ++i) {
//             long v = 1;
//             for (int j = 0; j < n; ++j) {
//                 if ((i >> j & 1) == 1) {
//                     v = lcm(v, coins[j]);
//                     if (v > mx) {
//                         break;
//                     }
//                 }
//             }
//             int m = Integer.bitCount(i);
//             if (m % 2 == 1) {
//                 cnt += mx / v;
//             } else {
//                 cnt -= mx / v;
//             }
//         }
//         return cnt >= k;
//     }
// 
//     private long lcm(long a, long b) {
//         return a * b / gcd(a, b);
//     }
// 
//     private long gcd(long a, long b) {
//         return b == 0 ? a : gcd(b, a % b);
//     }
// }

// Accepted solution for LeetCode #3116: Kth Smallest Amount With Single Denomination Combination
function findKthSmallest(coins: number[], k: number): number {
    let [l, r] = [1n, BigInt(1e11)];
    const n = coins.length;
    const check = (mx: bigint): boolean => {
        let cnt = 0n;
        for (let i = 1; i < 1 << n; ++i) {
            let v = 1n;
            for (let j = 0; j < n; ++j) {
                if ((i >> j) & 1) {
                    v = lcm(v, BigInt(coins[j]));
                    if (v > mx) {
                        break;
                    }
                }
            }
            const m = bitCount(i);
            if (m & 1) {
                cnt += mx / v;
            } else {
                cnt -= mx / v;
            }
        }
        return cnt >= BigInt(k);
    };
    while (l < r) {
        const mid = (l + r) >> 1n;
        if (check(mid)) {
            r = mid;
        } else {
            l = mid + 1n;
        }
    }
    return Number(l);
}

function gcd(a: bigint, b: bigint): bigint {
    return b === 0n ? a : gcd(b, a % b);
}

function lcm(a: bigint, b: bigint): bigint {
    return (a * b) / gcd(a, b);
}

function bitCount(i: number): number {
    i = i - ((i >>> 1) & 0x55555555);
    i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
    i = (i + (i >>> 4)) & 0x0f0f0f0f;
    i = i + (i >>> 8);
    i = i + (i >>> 16);
    return i & 0x3f;
}

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.