LeetCode #3471 — EASY

Find the Largest Almost Missing Integer

Build confidence with an intuition-first walkthrough focused on array fundamentals.

Solve on LeetCode

The Problem

Problem Statement

You are given an integer array nums and an integer k.

An integer x is almost missing from nums if x appears in exactly one subarray of size k within nums.

Return the largest almost missing integer from nums. If no such integer exists, return -1.

A subarray is a contiguous sequence of elements within an array.

Example 1:

Input: nums = [3,9,2,1,7], k = 3

Output: 7

Explanation:

1 appears in 2 subarrays of size 3: [9, 2, 1] and [2, 1, 7].
2 appears in 3 subarrays of size 3: [3, 9, 2], [9, 2, 1], [2, 1, 7].
3 appears in 1 subarray of size 3: [3, 9, 2].
7 appears in 1 subarray of size 3: [2, 1, 7].
9 appears in 2 subarrays of size 3: [3, 9, 2], and [9, 2, 1].

We return 7 since it is the largest integer that appears in exactly one subarray of size k.

Example 2:

Input: nums = [3,9,7,2,1,7], k = 4

Output: 3

Explanation:

1 appears in 2 subarrays of size 4: [9, 7, 2, 1], [7, 2, 1, 7].
2 appears in 3 subarrays of size 4: [3, 9, 7, 2], [9, 7, 2, 1], [7, 2, 1, 7].
3 appears in 1 subarray of size 4: [3, 9, 7, 2].
7 appears in 3 subarrays of size 4: [3, 9, 7, 2], [9, 7, 2, 1], [7, 2, 1, 7].
9 appears in 2 subarrays of size 4: [3, 9, 7, 2], [9, 7, 2, 1].

We return 3 since it is the largest and only integer that appears in exactly one subarray of size k.

Example 3:

Input: nums = [0,0], k = 1

Output: -1

Explanation:

There is no integer that appears in only one subarray of size 1.

Constraints:

1 <= nums.length <= 50
0 <= nums[i] <= 50
1 <= k <= nums.length

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums and an integer k. An integer x is almost missing from nums if x appears in exactly one subarray of size k within nums. Return the largest almost missing integer from nums. If no such integer exists, return -1. A subarray is a contiguous sequence of elements within an array.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

[3,9,2,1,7]
3

Example 2

[3,9,7,2,1,7]
4

Example 3

[0,0]
1

Core Insight

What unlocks the optimal approach

Solve the problem for three different cases: <code>k = 1</code>, <code>k = n</code>, and <code>1 < k < n</code>
If <code>k = 1</code>, return the largest element that occurs exactly once in <code>nums</code>
If <code>k = n</code>, return the largest element in <code>nums</code>
If <code>1 < k < n</code>, all elements different from <code>nums[0]</code> and <code>nums[n - 1]</code> will occur in more than one subarray of size <code>k</code>. Hence, the answer is the largest of <code>nums[0]</code> and <code>nums[n - 1]</code> if they both occur exactly once in the array. If one of them occurs more than once, return the other. If both of them occur more than once, return -1.

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 #3471: Find the Largest Almost Missing Integer
class Solution {
    private int[] nums;

    public int largestInteger(int[] nums, int k) {
        this.nums = nums;
        if (k == 1) {
            Map<Integer, Integer> cnt = new HashMap<>();
            for (int x : nums) {
                cnt.merge(x, 1, Integer::sum);
            }
            int ans = -1;
            for (var e : cnt.entrySet()) {
                if (e.getValue() == 1) {
                    ans = Math.max(ans, e.getKey());
                }
            }
            return ans;
        }
        if (k == nums.length) {
            return Arrays.stream(nums).max().getAsInt();
        }
        return Math.max(f(0), f(nums.length - 1));
    }

    private int f(int k) {
        for (int i = 0; i < nums.length; ++i) {
            if (i != k && nums[i] == nums[k]) {
                return -1;
            }
        }
        return nums[k];
    }
}

// Accepted solution for LeetCode #3471: Find the Largest Almost Missing Integer
func largestInteger(nums []int, k int) int {
    if k == 1 {
        cnt := make(map[int]int)
        for _, x := range nums {
            cnt[x]++
        }
        ans := -1
        for x, v := range cnt {
            if v == 1 {
                ans = max(ans, x)
            }
        }
        return ans
    }

    n := len(nums)
    if k == n {
        return slices.Max(nums)
    }

    f := func(k int) int {
        for i, x := range nums {
            if i != k && x == nums[k] {
                return -1
            }
        }
        return nums[k]
    }

    return max(f(0), f(n-1))
}

# Accepted solution for LeetCode #3471: Find the Largest Almost Missing Integer
class Solution:
    def largestInteger(self, nums: List[int], k: int) -> int:
        def f(k: int) -> int:
            for i, x in enumerate(nums):
                if i != k and x == nums[k]:
                    return -1
            return nums[k]

        if k == 1:
            cnt = Counter(nums)
            return max((x for x, v in cnt.items() if v == 1), default=-1)
        if k == len(nums):
            return max(nums)
        return max(f(0), f(len(nums) - 1))

// Accepted solution for LeetCode #3471: Find the Largest Almost Missing Integer
// 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 #3471: Find the Largest Almost Missing Integer
// class Solution {
//     private int[] nums;
// 
//     public int largestInteger(int[] nums, int k) {
//         this.nums = nums;
//         if (k == 1) {
//             Map<Integer, Integer> cnt = new HashMap<>();
//             for (int x : nums) {
//                 cnt.merge(x, 1, Integer::sum);
//             }
//             int ans = -1;
//             for (var e : cnt.entrySet()) {
//                 if (e.getValue() == 1) {
//                     ans = Math.max(ans, e.getKey());
//                 }
//             }
//             return ans;
//         }
//         if (k == nums.length) {
//             return Arrays.stream(nums).max().getAsInt();
//         }
//         return Math.max(f(0), f(nums.length - 1));
//     }
// 
//     private int f(int k) {
//         for (int i = 0; i < nums.length; ++i) {
//             if (i != k && nums[i] == nums[k]) {
//                 return -1;
//             }
//         }
//         return nums[k];
//     }
// }

// Accepted solution for LeetCode #3471: Find the Largest Almost Missing Integer
function largestInteger(nums: number[], k: number): number {
    if (k === 1) {
        const cnt = new Map<number, number>();
        for (const x of nums) {
            cnt.set(x, (cnt.get(x) || 0) + 1);
        }
        let ans = -1;
        for (const [x, v] of cnt.entries()) {
            if (v === 1 && x > ans) {
                ans = x;
            }
        }
        return ans;
    }

    const n = nums.length;
    if (k === n) {
        return Math.max(...nums);
    }

    const f = (k: number): number => {
        for (let i = 0; i < n; i++) {
            if (i !== k && nums[i] === nums[k]) {
                return -1;
            }
        }
        return nums[k];
    };

    return Math.max(f(0), f(n - 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(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.

Find the Largest Almost Missing Integer

Problem Statement

Roadmap

Brute Force Baseline

Baseline thinking

Example 1

Example 2

Example 3

Related Problems

Core Insight

What unlocks the optimal approach

Algorithm Walkthrough

Iteration Checklist

Edge Cases

Full Annotated Code

Interactive Study Demo

Complexity Analysis

Approach Breakdown

Common Mistakes

Off-by-one on range boundaries

Mutating counts without cleanup