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Problem solution · Python

Find X Value of Array II

Find X Value of Array II: a Python solution using segment tree or range structure. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Segment tree or range structure
Source
Kamyu LeetCode Solutions
Length
82 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Segment tree or range structure

For Find X Value of Array II, the implementation stores interval information in a range-query data structure so updates and queries avoid rescanning the full input.

  1. Choose the aggregate stored for each interval or prefix.
  2. Build or initialize the structure from the input.
  3. Apply updates and combine the affected nodes to answer each query.

Code notes

  • 82 lines of Python from the credited upstream file find-x-value-of-array-ii.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Count the build once, then multiply the logarithmic update or query path by the number of operations.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeFind X Value of Array II · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(n * k + q * k * logn)# Space: O(n * k) # segment treeclass Solution(object):    def resultArray(self, nums, k, queries):        """        :type nums: List[int]        :type k: int        :type queries: List[List[int]]        :rtype: List[int]        """        # Template:        # https://github.com/kamyu104/LeetCode-Solutions/blob/master/Python/block-placement-queries.py        class SegmentTree(object):            def __init__(self, N,                         build_fn=lambda _: None,                         query_fn=lambda x, y: y if x is None else x if y is None else max(x, y),                         update_fn=lambda x: x):                self.tree = [None]*(1<<((N-1).bit_length()+1))                self.base = len(self.tree)>>1                self.query_fn = query_fn                self.update_fn = update_fn                for i in xrange(self.base, self.base+N):                    self.tree[i] = build_fn(i-self.base)                for i in reversed(xrange(1, self.base)):                    self.tree[i] = query_fn(self.tree[i<<1], self.tree[(i<<1)+1])             def update(self, i, h):                x = self.base+i                self.tree[x] = self.update_fn(h)                while x > 1:                    x >>= 1                    self.tree[x] = self.query_fn(self.tree[x<<1], self.tree[(x<<1)+1])             def query(self, L, R):                L += self.base                R += self.base                left = right = None                while L <= R:                    if L & 1:                        left = self.query_fn(left, self.tree[L])                        L += 1                    if R & 1 == 0:                        right = self.query_fn(self.tree[R], right)                        R -= 1                    L >>= 1                    R >>= 1                return self.query_fn(left, right)                def build(i):            x = nums[i]%k            cnt = [0]*(k+1)            cnt[x] = 1            cnt[-1] = x            return cnt         def update(x):            x %= k            cnt = [0]*(k+1)            cnt[x] = 1            cnt[-1] = x            return cnt                    def query(x, y):            if x is None:                return y            if y is None:                return x            cnt = x[:]            for i in xrange(k):                cnt[x[-1]*i%k] += y[i]            cnt[-1] = x[-1]*y[-1]%k            return cnt                st = SegmentTree(len(nums), build_fn=build, update_fn=update, query_fn=query)        result = [0]*len(queries)        for idx, (i, v, s, x) in enumerate(queries):            st.update(i, v)            result[idx] = st.query(s, len(nums)-1)[x]        return result

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