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3471. Find the Largest Almost Missing Integer

Description

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

Solutions

Solution 1: Case Analysis

If k=1, then each element in the array forms a subarray of size 1. In this case, we only need to find the maximum value among the elements that appear exactly once in the array.

If k=n, then the entire array forms a subarray of size n. In this case, we only need to return the maximum value in the array.

If 1<k<n, only nums[0] and nums[n−1] can be the almost missing integers. If they appear elsewhere in the array, they are not almost missing integers. Therefore, we only need to check if nums[0] and nums[n−1] appear elsewhere in the array and return the maximum value among them.

If no almost missing integer exists, return −1.

The time complexity is O(n), and the space complexity is O(n). Here, n is the length of the array nums.

  • 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];
    }
}


  • class Solution {
public:
    int largestInteger(vector<int>& nums, int k) {
        if (k == 1) {
            unordered_map<int, int> cnt;
            for (int x : nums) {
                ++cnt[x];
            }
            int ans = -1;
            for (auto& [x, v] : cnt) {
                if (v == 1) {
                    ans = max(ans, x);
                }
            }
            return ans;
        }
        int n = nums.size();
        if (k == n) {
            return ranges::max(nums);
        }
        auto f = [&](int k) -> int {
            for (int i = 0; i < n; ++i) {
                if (i != k && nums[i] == nums[k]) {
                    return -1;
                }
            }
            return nums[k];
        };
        return max(f(0), f(n - 1));
    }
};


  • 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))


  • 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))
}


  • 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));
}

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