LeetCode #2518 — HARD

Number of Great Partitions

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

The Problem

Problem Statement

You are given an array nums consisting of positive integers and an integer k.

Partition the array into two ordered groups such that each element is in exactly one group. A partition is called great if the sum of elements of each group is greater than or equal to k.

Return the number of distinct great partitions. Since the answer may be too large, return it modulo 10⁹ + 7.

Two partitions are considered distinct if some element nums[i] is in different groups in the two partitions.

Example 1:

Input: nums = [1,2,3,4], k = 4
Output: 6
Explanation: The great partitions are: ([1,2,3], [4]), ([1,3], [2,4]), ([1,4], [2,3]), ([2,3], [1,4]), ([2,4], [1,3]) and ([4], [1,2,3]).

Example 2:

Input: nums = [3,3,3], k = 4
Output: 0
Explanation: There are no great partitions for this array.

Example 3:

Input: nums = [6,6], k = 2
Output: 2
Explanation: We can either put nums[0] in the first partition or in the second partition.
The great partitions will be ([6], [6]) and ([6], [6]).

Constraints:

1 <= nums.length, k <= 1000
1 <= nums[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given an array nums consisting of positive integers and an integer k. Partition the array into two ordered groups such that each element is in exactly one group. A partition is called great if the sum of elements of each group is greater than or equal to k. Return the number of distinct great partitions. Since the answer may be too large, return it modulo 109 + 7. Two partitions are considered distinct if some element nums[i] is in different groups in the two partitions.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming

Example 1

[1,2,3,4]
4

Example 2

[3,3,3]
4

Example 3

[6,6]
2

Core Insight

What unlocks the optimal approach

If the sum of the array is smaller than 2*k, then it is impossible to find a great partition.
Solve the reverse problem, that is, find the number of partitions where the sum of elements of at least one of the two groups is smaller than k.

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 #2518: Number of Great Partitions
class Solution {
    private static final int MOD = (int) 1e9 + 7;

    public int countPartitions(int[] nums, int k) {
        long s = 0;
        for (int v : nums) {
            s += v;
        }
        if (s < k * 2) {
            return 0;
        }
        int n = nums.length;
        long[][] f = new long[n + 1][k];
        f[0][0] = 1;
        long ans = 1;
        for (int i = 1; i <= n; ++i) {
            int v = nums[i - 1];
            ans = ans * 2 % MOD;
            for (int j = 0; j < k; ++j) {
                f[i][j] = f[i - 1][j];
                if (j >= v) {
                    f[i][j] = (f[i][j] + f[i - 1][j - v]) % MOD;
                }
            }
        }
        for (int j = 0; j < k; ++j) {
            ans = (ans - f[n][j] * 2 % MOD + MOD) % MOD;
        }
        return (int) ans;
    }
}

// Accepted solution for LeetCode #2518: Number of Great Partitions
func countPartitions(nums []int, k int) int {
	s := 0
	for _, v := range nums {
		s += v
	}
	if s < k*2 {
		return 0
	}
	const mod int = 1e9 + 7
	n := len(nums)
	f := make([][]int, n+1)
	for i := range f {
		f[i] = make([]int, k)
	}
	f[0][0] = 1
	ans := 1
	for i := 1; i <= n; i++ {
		v := nums[i-1]
		ans = ans * 2 % mod
		for j := 0; j < k; j++ {
			f[i][j] = f[i-1][j]
			if j >= v {
				f[i][j] = (f[i][j] + f[i-1][j-v]) % mod
			}
		}
	}
	for j := 0; j < k; j++ {
		ans = (ans - f[n][j]*2%mod + mod) % mod
	}
	return ans
}

# Accepted solution for LeetCode #2518: Number of Great Partitions
class Solution:
    def countPartitions(self, nums: List[int], k: int) -> int:
        if sum(nums) < k * 2:
            return 0
        mod = 10**9 + 7
        n = len(nums)
        f = [[0] * k for _ in range(n + 1)]
        f[0][0] = 1
        ans = 1
        for i in range(1, n + 1):
            ans = ans * 2 % mod
            for j in range(k):
                f[i][j] = f[i - 1][j]
                if j >= nums[i - 1]:
                    f[i][j] = (f[i][j] + f[i - 1][j - nums[i - 1]]) % mod
        return (ans - sum(f[-1]) * 2 + mod) % mod

// Accepted solution for LeetCode #2518: Number of Great Partitions
// 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 #2518: Number of Great Partitions
// class Solution {
//     private static final int MOD = (int) 1e9 + 7;
// 
//     public int countPartitions(int[] nums, int k) {
//         long s = 0;
//         for (int v : nums) {
//             s += v;
//         }
//         if (s < k * 2) {
//             return 0;
//         }
//         int n = nums.length;
//         long[][] f = new long[n + 1][k];
//         f[0][0] = 1;
//         long ans = 1;
//         for (int i = 1; i <= n; ++i) {
//             int v = nums[i - 1];
//             ans = ans * 2 % MOD;
//             for (int j = 0; j < k; ++j) {
//                 f[i][j] = f[i - 1][j];
//                 if (j >= v) {
//                     f[i][j] = (f[i][j] + f[i - 1][j - v]) % MOD;
//                 }
//             }
//         }
//         for (int j = 0; j < k; ++j) {
//             ans = (ans - f[n][j] * 2 % MOD + MOD) % MOD;
//         }
//         return (int) ans;
//     }
// }

// Accepted solution for LeetCode #2518: Number of Great Partitions
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2518: Number of Great Partitions
// class Solution {
//     private static final int MOD = (int) 1e9 + 7;
// 
//     public int countPartitions(int[] nums, int k) {
//         long s = 0;
//         for (int v : nums) {
//             s += v;
//         }
//         if (s < k * 2) {
//             return 0;
//         }
//         int n = nums.length;
//         long[][] f = new long[n + 1][k];
//         f[0][0] = 1;
//         long ans = 1;
//         for (int i = 1; i <= n; ++i) {
//             int v = nums[i - 1];
//             ans = ans * 2 % MOD;
//             for (int j = 0; j < k; ++j) {
//                 f[i][j] = f[i - 1][j];
//                 if (j >= v) {
//                     f[i][j] = (f[i][j] + f[i - 1][j - v]) % MOD;
//                 }
//             }
//         }
//         for (int j = 0; j < k; ++j) {
//             ans = (ans - f[n][j] * 2 % MOD + MOD) % MOD;
//         }
//         return (int) 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(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

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.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.