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2216. Minimum Deletions to Make Array Beautiful

Description

You are given a 0-indexed integer array nums. The array nums is beautiful if:

  • nums.length is even.
  • nums[i] != nums[i + 1] for all i % 2 == 0.

Note that an empty array is considered beautiful.

You can delete any number of elements from nums. When you delete an element, all the elements to the right of the deleted element will be shifted one unit to the left to fill the gap created and all the elements to the left of the deleted element will remain unchanged.

Return the minimum number of elements to delete from nums to make it beautiful.

 

Example 1:

Input: nums = [1,1,2,3,5]
Output: 1
Explanation: You can delete either nums[0] or nums[1] to make nums = [1,2,3,5] which is beautiful. It can be proven you need at least 1 deletion to make nums beautiful.

Example 2:

Input: nums = [1,1,2,2,3,3]
Output: 2
Explanation: You can delete nums[0] and nums[5] to make nums = [1,2,2,3] which is beautiful. It can be proven you need at least 2 deletions to make nums beautiful.

 

Constraints:

  • 1 <= nums.length <= 105
  • 0 <= nums[i] <= 105

Solutions

Solution 1: Greedy

Thinking

A beautiful array has even length and \(nums[2k] \neq nums[2k+1]\) for every pair. \(n \le 10^5\) forbids searching deletion subsets. Pairing left to right, a pair with \(nums[i]=nums[i+1]\) cannot be kept, so one element must go.

Delete the current \(nums[i]\) (advance one step and count a deletion); if the two values differ, keep the pair and skip two indices. If the leftover length is odd, delete one more at the end.

According to the problem description, we know that a beautiful array has an even number of elements, and if we divide every two adjacent elements in this array into a group, then the two elements in each group are not equal. This means that the elements within a group cannot be repeated, but the elements between groups can be repeated.

Therefore, we consider traversing the array from left to right. As long as we encounter two adjacent elements that are equal, we delete one of them, that is, the deletion count increases by one; otherwise, we can keep these two elements.

Finally, we check whether the length of the array after deletion is even. If not, it means that we need to delete one more element to make the final array length even.

The time complexity is \(O(n)\), where \(n\) is the length of the array. We only need to traverse the array once. The space complexity is \(O(1)\).

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class Solution:
    def minDeletion(self, nums: List[int]) -> int:
        n = len(nums)
        i = ans = 0
        while i < n - 1:
            if nums[i] == nums[i + 1]:
                ans += 1
                i += 1
            else:
                i += 2
        ans += (n - ans) % 2
        return ans

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class Solution {
    public int minDeletion(int[] nums) {
        int n = nums.length;
        int ans = 0;
        for (int i = 0; i < n - 1; ++i) {
            if (nums[i] == nums[i + 1]) {
                ++ans;
            } else {
                ++i;
            }
        }
        ans += (n - ans) % 2;
        return ans;
    }
}

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class Solution {
public:
    int minDeletion(vector<int>& nums) {
        int n = nums.size();
        int ans = 0;
        for (int i = 0; i < n - 1; ++i) {
            if (nums[i] == nums[i + 1]) {
                ++ans;
            } else {
                ++i;
            }
        }
        ans += (n - ans) % 2;
        return ans;
    }
};

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func minDeletion(nums []int) (ans int) {
    n := len(nums)
    for i := 0; i < n-1; i++ {
        if nums[i] == nums[i+1] {
            ans++
        } else {
            i++
        }
    }
    ans += (n - ans) % 2
    return
}

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function minDeletion(nums: number[]): number {
    const n = nums.length;
    let ans = 0;
    for (let i = 0; i < n - 1; ++i) {
        if (nums[i] === nums[i + 1]) {
            ++ans;
        } else {
            ++i;
        }
    }
    ans += (n - ans) % 2;
    return ans;
}

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impl Solution {
    pub fn min_deletion(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let mut ans = 0;
        let mut i = 0;
        while i < n - 1 {
            if nums[i] == nums[i + 1] {
                ans += 1;
                i += 1;
            } else {
                i += 2;
            }
        }
        ans += (n - ans) % 2;
        ans as i32
    }
}

Solution 2

Thinking

Solution 1 decides deletion on each adjacent pair. We can instead collapse a run of equal values: extras inside the run must be deleted, one copy stays as the even index, and it pairs with the next different value.

The inner loop skips equals and counts those deletions, then the pointer jumps past the paired partner. The same parity cleanup applies. Still \(O(n)\), advancing by equal runs.

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class Solution:
    def minDeletion(self, nums: List[int]) -> int:
        n = len(nums)
        ans = i = 0
        while i < n:
            j = i + 1
            while j < n and nums[j] == nums[i]:
                j += 1
                ans += 1
            i = j + 1
        ans += (n - ans) % 2
        return ans

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class Solution {
    public int minDeletion(int[] nums) {
        int n = nums.length;
        int ans = 0;
        for (int i = 0; i < n;) {
            int j = i + 1;
            while (j < n && nums[j] == nums[i]) {
                ++j;
                ++ans;
            }
            i = j + 1;
        }
        ans += (n - ans) % 2;
        return ans;
    }
}

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class Solution {
public:
    int minDeletion(vector<int>& nums) {
        int n = nums.size();
        int ans = 0;
        for (int i = 0; i < n;) {
            int j = i + 1;
            while (j < n && nums[j] == nums[i]) {
                ++j;
                ++ans;
            }
            i = j + 1;
        }
        ans += (n - ans) % 2;
        return ans;
    }
};

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func minDeletion(nums []int) (ans int) {
    n := len(nums)
    for i := 0; i < n; {
        j := i + 1
        for ; j < n && nums[j] == nums[i]; j++ {
            ans++
        }
        i = j + 1
    }
    ans += (n - ans) % 2
    return
}

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function minDeletion(nums: number[]): number {
    const n = nums.length;
    let ans = 0;
    for (let i = 0; i < n;) {
        let j = i + 1;
        for (; j < n && nums[j] === nums[i]; ++j) {
            ++ans;
        }
        i = j + 1;
    }
    ans += (n - ans) % 2;
    return ans;
}

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impl Solution {
    pub fn min_deletion(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let mut ans = 0;
        let mut i = 0;
        while i < n {
            let mut j = i + 1;
            while j < n && nums[j] == nums[i] {
                ans += 1;
                j += 1;
            }
            i = j + 1;
        }
        ans += (n - ans) % 2;
        ans as i32
    }
}

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