LeetCode #2780 — MEDIUM

Minimum Index of a Valid Split

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

The Problem

Problem Statement

An element x of an integer array arr of length m is dominant if more than half the elements of arr have a value of x.

You are given a 0-indexed integer array nums of length n with one dominant element.

You can split nums at an index i into two arrays nums[0, ..., i] and nums[i + 1, ..., n - 1], but the split is only valid if:

0 <= i < n - 1
nums[0, ..., i], and nums[i + 1, ..., n - 1] have the same dominant element.

Here, nums[i, ..., j] denotes the subarray of nums starting at index i and ending at index j, both ends being inclusive. Particularly, if j < i then nums[i, ..., j] denotes an empty subarray.

Return the minimum index of a valid split. If no valid split exists, return -1.

Example 1:

Input: nums = [1,2,2,2]
Output: 2
Explanation: We can split the array at index 2 to obtain arrays [1,2,2] and [2]. 
In array [1,2,2], element 2 is dominant since it occurs twice in the array and 2 * 2 > 3. 
In array [2], element 2 is dominant since it occurs once in the array and 1 * 2 > 1.
Both [1,2,2] and [2] have the same dominant element as nums, so this is a valid split. 
It can be shown that index 2 is the minimum index of a valid split.

Example 2:

Input: nums = [2,1,3,1,1,1,7,1,2,1]
Output: 4
Explanation: We can split the array at index 4 to obtain arrays [2,1,3,1,1] and [1,7,1,2,1].
In array [2,1,3,1,1], element 1 is dominant since it occurs thrice in the array and 3 * 2 > 5.
In array [1,7,1,2,1], element 1 is dominant since it occurs thrice in the array and 3 * 2 > 5.
Both [2,1,3,1,1] and [1,7,1,2,1] have the same dominant element as nums, so this is a valid split.
It can be shown that index 4 is the minimum index of a valid split.

Example 3:

Input: nums = [3,3,3,3,7,2,2]
Output: -1
Explanation: It can be shown that there is no valid split.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹
nums has exactly one dominant element.

Step 01

Brute Force Baseline

Problem summary: An element x of an integer array arr of length m is dominant if more than half the elements of arr have a value of x. You are given a 0-indexed integer array nums of length n with one dominant element. You can split nums at an index i into two arrays nums[0, ..., i] and nums[i + 1, ..., n - 1], but the split is only valid if: 0 <= i < n - 1 nums[0, ..., i], and nums[i + 1, ..., n - 1] have the same dominant element. Here, nums[i, ..., j] denotes the subarray of nums starting at index i and ending at index j, both ends being inclusive. Particularly, if j < i then nums[i, ..., j] denotes an empty subarray. Return the minimum index of a valid split. If no valid split exists, return -1.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

[1,2,2,2]

Example 2

[2,1,3,1,1,1,7,1,2,1]

Example 3

[3,3,3,3,7,2,2]

Core Insight

What unlocks the optimal approach

Find the dominant element of nums by using a hashmap to maintain element frequency, we denote the dominant element as x and its frequency as f.
For each index in [0, n - 2], calculate f1, x’s frequency in the subarray [0, i] when looping the index. And f2, x’s frequency in the subarray [i + 1, n - 1] which is equal to f - f1. Then we can check whether x is dominant in both subarrays.

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 #2780: Minimum Index of a Valid Split
class Solution {
    public int minimumIndex(List<Integer> nums) {
        int x = 0, cnt = 0;
        Map<Integer, Integer> freq = new HashMap<>();
        for (int v : nums) {
            int t = freq.merge(v, 1, Integer::sum);
            if (cnt < t) {
                cnt = t;
                x = v;
            }
        }
        int cur = 0;
        for (int i = 1; i <= nums.size(); ++i) {
            if (nums.get(i - 1) == x) {
                ++cur;
                if (cur * 2 > i && (cnt - cur) * 2 > nums.size() - i) {
                    return i - 1;
                }
            }
        }
        return -1;
    }
}

// Accepted solution for LeetCode #2780: Minimum Index of a Valid Split
func minimumIndex(nums []int) int {
	x, cnt := 0, 0
	freq := map[int]int{}
	for _, v := range nums {
		freq[v]++
		if freq[v] > cnt {
			x, cnt = v, freq[v]
		}
	}
	cur := 0
	for i, v := range nums {
		i++
		if v == x {
			cur++
			if cur*2 > i && (cnt-cur)*2 > len(nums)-i {
				return i - 1
			}
		}
	}
	return -1
}

# Accepted solution for LeetCode #2780: Minimum Index of a Valid Split
class Solution:
    def minimumIndex(self, nums: List[int]) -> int:
        x, cnt = Counter(nums).most_common(1)[0]
        cur = 0
        for i, v in enumerate(nums, 1):
            if v == x:
                cur += 1
                if cur * 2 > i and (cnt - cur) * 2 > len(nums) - i:
                    return i - 1
        return -1

// Accepted solution for LeetCode #2780: Minimum Index of a Valid Split
// 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 #2780: Minimum Index of a Valid Split
// class Solution {
//     public int minimumIndex(List<Integer> nums) {
//         int x = 0, cnt = 0;
//         Map<Integer, Integer> freq = new HashMap<>();
//         for (int v : nums) {
//             int t = freq.merge(v, 1, Integer::sum);
//             if (cnt < t) {
//                 cnt = t;
//                 x = v;
//             }
//         }
//         int cur = 0;
//         for (int i = 1; i <= nums.size(); ++i) {
//             if (nums.get(i - 1) == x) {
//                 ++cur;
//                 if (cur * 2 > i && (cnt - cur) * 2 > nums.size() - i) {
//                     return i - 1;
//                 }
//             }
//         }
//         return -1;
//     }
// }

// Accepted solution for LeetCode #2780: Minimum Index of a Valid Split
function minimumIndex(nums: number[]): number {
    let [x, cnt] = [0, 0];
    const freq: Map<number, number> = new Map();
    for (const v of nums) {
        freq.set(v, (freq.get(v) ?? 0) + 1);
        if (freq.get(v)! > cnt) {
            [x, cnt] = [v, freq.get(v)!];
        }
    }
    let cur = 0;
    for (let i = 1; i <= nums.length; ++i) {
        if (nums[i - 1] === x) {
            ++cur;
            if (cur * 2 > i && (cnt - cur) * 2 > nums.length - i) {
                return i - 1;
            }
        }
    }
    return -1;
}

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.

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.