LeetCode #2963 — HARD

Count the Number of Good Partitions

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

The Problem

Problem Statement

You are given a 0-indexed array nums consisting of positive integers.

A partition of an array into one or more contiguous subarrays is called good if no two subarrays contain the same number.

Return the total number of good partitions of nums.

Since the answer may be large, return it modulo 10⁹ + 7.

Example 1:

Input: nums = [1,2,3,4]
Output: 8
Explanation: The 8 possible good partitions are: ([1], [2], [3], [4]), ([1], [2], [3,4]), ([1], [2,3], [4]), ([1], [2,3,4]), ([1,2], [3], [4]), ([1,2], [3,4]), ([1,2,3], [4]), and ([1,2,3,4]).

Example 2:

Input: nums = [1,1,1,1]
Output: 1
Explanation: The only possible good partition is: ([1,1,1,1]).

Example 3:

Input: nums = [1,2,1,3]
Output: 2
Explanation: The 2 possible good partitions are: ([1,2,1], [3]) and ([1,2,1,3]).

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed array nums consisting of positive integers. A partition of an array into one or more contiguous subarrays is called good if no two subarrays contain the same number. Return the total number of good partitions of nums. Since the answer may be large, return it modulo 109 + 7.

Baseline thinking

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

Pattern signal: Array · Hash Map · Math

Example 1

[1,2,3,4]

Example 2

[1,1,1,1]

Example 3

[1,2,1,3]

Core Insight

What unlocks the optimal approach

If a segment contains a value, it must contain all occurrences of the same value.
Partition the array into segments making each one as short as possible. This can be achieved by two-pointers or using a Set.
If we have <code>m</code> segments, we can arbitrarily group the neighboring segments. How many ways are there to group these <code>m</code> segments?

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 #2963: Count the Number of Good Partitions
class Solution {
    public int numberOfGoodPartitions(int[] nums) {
        Map<Integer, Integer> last = new HashMap<>();
        int n = nums.length;
        for (int i = 0; i < n; ++i) {
            last.put(nums[i], i);
        }
        final int mod = (int) 1e9 + 7;
        int j = -1;
        int k = 0;
        for (int i = 0; i < n; ++i) {
            j = Math.max(j, last.get(nums[i]));
            k += i == j ? 1 : 0;
        }
        return qpow(2, k - 1, mod);
    }

    private int qpow(long a, int n, int mod) {
        long ans = 1;
        for (; n > 0; n >>= 1) {
            if ((n & 1) == 1) {
                ans = ans * a % mod;
            }
            a = a * a % mod;
        }
        return (int) ans;
    }
}

// Accepted solution for LeetCode #2963: Count the Number of Good Partitions
func numberOfGoodPartitions(nums []int) int {
	qpow := func(a, n, mod int) int {
		ans := 1
		for ; n > 0; n >>= 1 {
			if n&1 == 1 {
				ans = ans * a % mod
			}
			a = a * a % mod
		}
		return ans
	}
	last := map[int]int{}
	for i, x := range nums {
		last[x] = i
	}
	const mod int = 1e9 + 7
	j, k := -1, 0
	for i, x := range nums {
		j = max(j, last[x])
		if i == j {
			k++
		}
	}
	return qpow(2, k-1, mod)
}

# Accepted solution for LeetCode #2963: Count the Number of Good Partitions
class Solution:
    def numberOfGoodPartitions(self, nums: List[int]) -> int:
        last = {x: i for i, x in enumerate(nums)}
        mod = 10**9 + 7
        j, k = -1, 0
        for i, x in enumerate(nums):
            j = max(j, last[x])
            k += i == j
        return pow(2, k - 1, mod)

// Accepted solution for LeetCode #2963: Count the Number of Good 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 #2963: Count the Number of Good Partitions
// class Solution {
//     public int numberOfGoodPartitions(int[] nums) {
//         Map<Integer, Integer> last = new HashMap<>();
//         int n = nums.length;
//         for (int i = 0; i < n; ++i) {
//             last.put(nums[i], i);
//         }
//         final int mod = (int) 1e9 + 7;
//         int j = -1;
//         int k = 0;
//         for (int i = 0; i < n; ++i) {
//             j = Math.max(j, last.get(nums[i]));
//             k += i == j ? 1 : 0;
//         }
//         return qpow(2, k - 1, mod);
//     }
// 
//     private int qpow(long a, int n, int mod) {
//         long ans = 1;
//         for (; n > 0; n >>= 1) {
//             if ((n & 1) == 1) {
//                 ans = ans * a % mod;
//             }
//             a = a * a % mod;
//         }
//         return (int) ans;
//     }
// }

// Accepted solution for LeetCode #2963: Count the Number of Good Partitions
function numberOfGoodPartitions(nums: number[]): number {
    const qpow = (a: number, n: number, mod: number) => {
        let ans = 1;
        for (; n; n >>= 1) {
            if (n & 1) {
                ans = Number((BigInt(ans) * BigInt(a)) % BigInt(mod));
            }
            a = Number((BigInt(a) * BigInt(a)) % BigInt(mod));
        }
        return ans;
    };
    const last: Map<number, number> = new Map();
    const n = nums.length;
    for (let i = 0; i < n; ++i) {
        last.set(nums[i], i);
    }
    const mod = 1e9 + 7;
    let [j, k] = [-1, 0];
    for (let i = 0; i < n; ++i) {
        j = Math.max(j, last.get(nums[i])!);
        if (i === j) {
            ++k;
        }
    }
    return qpow(2, k - 1, mod);
}

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

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.

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.