1713. Minimum Operations to Make a Subsequence

Description

You are given an array target that consists of distinct integers and another integer array arr that can have duplicates.

In one operation, you can insert any integer at any position in arr. For example, if arr = [1,4,1,2], you can add 3 in the middle and make it [1,4,3,1,2]. Note that you can insert the integer at the very beginning or end of the array.

Return the minimum number of operations needed to make target a subsequence of arr.

A subsequence of an array is a new array generated from the original array by deleting some elements (possibly none) without changing the remaining elements' relative order. For example, [2,7,4] is a subsequence of [4,2,3,7,2,1,4] (the underlined elements), while [2,4,2] is not.

Example 1:

Input: target = [5,1,3], arr = [9,4,2,3,4]
Output: 2
Explanation: You can add 5 and 1 in such a way that makes arr = [5,9,4,1,2,3,4], then target will be a subsequence of arr.

Example 2:

Input: target = [6,4,8,1,3,2], arr = [4,7,6,2,3,8,6,1]
Output: 3

Constraints:

1 <= target.length, arr.length <= 10⁵
1 <= target[i], arr[i] <= 10⁹
target contains no duplicates.

Solutions

class BinaryIndexedTree {
    private int n;
    private int[] c;

    public BinaryIndexedTree(int n) {
        this.n = n;
        this.c = new int[n + 1];
    }

    public static int lowbit(int x) {
        return x & -x;
    }

    public void update(int x, int val) {
        while (x <= n) {
            c[x] = Math.max(c[x], val);
            x += lowbit(x);
        }
    }

    public int query(int x) {
        int s = 0;
        while (x > 0) {
            s = Math.max(s, c[x]);
            x -= lowbit(x);
        }
        return s;
    }
}

class Solution {
    public int minOperations(int[] target, int[] arr) {
        Map<Integer, Integer> d = new HashMap<>();
        for (int i = 0; i < target.length; ++i) {
            d.put(target[i], i);
        }
        List<Integer> nums = new ArrayList<>();
        for (int i = 0; i < arr.length; ++i) {
            if (d.containsKey(arr[i])) {
                nums.add(d.get(arr[i]));
            }
        }
        return target.length - lengthOfLIS(nums);
    }

    private int lengthOfLIS(List<Integer> nums) {
        TreeSet<Integer> ts = new TreeSet();
        for (int v : nums) {
            ts.add(v);
        }
        int idx = 1;
        Map<Integer, Integer> d = new HashMap<>();
        for (int v : ts) {
            d.put(v, idx++);
        }
        int ans = 0;
        BinaryIndexedTree tree = new BinaryIndexedTree(nums.size());
        for (int v : nums) {
            int x = d.get(v);
            int t = tree.query(x - 1) + 1;
            ans = Math.max(ans, t);
            tree.update(x, t);
        }
        return ans;
    }
}

class BinaryIndexedTree {
public:
    int n;
    vector<int> c;

    BinaryIndexedTree(int _n)
        : n(_n)
        , c(_n + 1) {}

    void update(int x, int val) {
        while (x <= n) {
            c[x] = max(c[x], val);
            x += lowbit(x);
        }
    }

    int query(int x) {
        int s = 0;
        while (x > 0) {
            s = max(s, c[x]);
            x -= lowbit(x);
        }
        return s;
    }

    int lowbit(int x) {
        return x & -x;
    }
};

class Solution {
public:
    int minOperations(vector<int>& target, vector<int>& arr) {
        unordered_map<int, int> d;
        for (int i = 0; i < target.size(); ++i) d[target[i]] = i;
        vector<int> nums;
        for (int i = 0; i < arr.size(); ++i) {
            if (d.count(arr[i])) {
                nums.push_back(d[arr[i]]);
            }
        }
        return target.size() - lengthOfLIS(nums);
    }

    int lengthOfLIS(vector<int>& nums) {
        set<int> s(nums.begin(), nums.end());
        int idx = 1;
        unordered_map<int, int> d;
        for (int v : s) d[v] = idx++;
        BinaryIndexedTree* tree = new BinaryIndexedTree(d.size());
        int ans = 0;
        for (int v : nums) {
            int x = d[v];
            int t = tree->query(x - 1) + 1;
            ans = max(ans, t);
            tree->update(x, t);
        }
        return ans;
    }
};

class BinaryIndexedTree:
    def __init__(self, n):
        self.n = n
        self.c = [0] * (n + 1)

    @staticmethod
    def lowbit(x):
        return x & -x

    def update(self, x, val):
        while x <= self.n:
            self.c[x] = max(self.c[x], val)
            x += BinaryIndexedTree.lowbit(x)

    def query(self, x):
        s = 0
        while x:
            s = max(s, self.c[x])
            x -= BinaryIndexedTree.lowbit(x)
        return s


class Solution:
    def minOperations(self, target: List[int], arr: List[int]) -> int:
        d = {v: i for i, v in enumerate(target)}
        nums = [d[v] for v in arr if v in d]
        return len(target) - self.lengthOfLIS(nums)

    def lengthOfLIS(self, nums):
        s = sorted(set(nums))
        m = {v: i for i, v in enumerate(s, 1)}
        tree = BinaryIndexedTree(len(m))
        ans = 0
        for v in nums:
            x = m[v]
            t = tree.query(x - 1) + 1
            ans = max(ans, t)
            tree.update(x, t)
        return ans

type BinaryIndexedTree struct {
	n int
	c []int
}

func newBinaryIndexedTree(n int) *BinaryIndexedTree {
	c := make([]int, n+1)
	return &BinaryIndexedTree{n, c}
}

func (this *BinaryIndexedTree) lowbit(x int) int {
	return x & -x
}

func (this *BinaryIndexedTree) update(x, val int) {
	for x <= this.n {
		if this.c[x] < val {
			this.c[x] = val
		}
		x += this.lowbit(x)
	}
}

func (this *BinaryIndexedTree) query(x int) int {
	s := 0
	for x > 0 {
		if s < this.c[x] {
			s = this.c[x]
		}
		x -= this.lowbit(x)
	}
	return s
}

func minOperations(target []int, arr []int) int {
	d := map[int]int{}
	for i, v := range target {
		d[v] = i
	}
	nums := []int{}
	for _, v := range arr {
		if i, ok := d[v]; ok {
			nums = append(nums, i)
		}
	}
	return len(target) - lengthOfLIS(nums)
}

func lengthOfLIS(nums []int) int {
	s := map[int]bool{}
	for _, v := range nums {
		s[v] = true
	}
	t := []int{}
	for v := range s {
		t = append(t, v)
	}
	sort.Ints(t)
	d := map[int]int{}
	for i, v := range t {
		d[v] = i + 1
	}
	tree := newBinaryIndexedTree(len(d))
	ans := 0
	for _, v := range nums {
		x := d[v]
		t := tree.query(x-1) + 1
		ans = max(ans, t)
		tree.update(x, t)
	}
	return ans
}

class BinaryIndexedTree {
    private n: number;
    private c: number[];

    constructor(n: number) {
        this.n = n;
        this.c = Array(n + 1).fill(0);
    }

    update(x: number, v: number): void {
        for (; x <= this.n; x += x & -x) {
            this.c[x] = Math.max(this.c[x], v);
        }
    }

    query(x: number): number {
        let ans = 0;
        for (; x > 0; x -= x & -x) {
            ans = Math.max(ans, this.c[x]);
        }
        return ans;
    }
}

function minOperations(target: number[], arr: number[]): number {
    const m = target.length;
    const d: Map<number, number> = new Map();
    target.forEach((x, i) => d.set(x, i + 1));
    const nums: number[] = [];
    arr.forEach(x => {
        if (d.has(x)) {
            nums.push(d.get(x)!);
        }
    });
    const tree = new BinaryIndexedTree(m);
    let ans = 0;
    nums.forEach(x => {
        const v = tree.query(x - 1) + 1;
        ans = Math.max(ans, v);
        tree.update(x, v);
    });
    return m - ans;
}

1713 - Minimum Operations to Make a Subsequence

1713. Minimum Operations to Make a Subsequence

Description

Solutions

All Problems

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1713. Minimum Operations to Make a Subsequence

Description

Solutions

All Problems

All Solutions