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**Restate the problem:** We need to find the power of all subarrays of size k. The power is the maximum element if the subarray contains consecutive integers in ascending order, otherwise -1.
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**1.1 Constraints & Complexity:**
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- Input size: `1 <= n == nums.length <= 500`, `1 <= nums[i] <= 10^5`, `1 <= k <= n`
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-**Time Complexity:** O(n * k) where n is the array length and k is the subarray size, as we check each of the (n-k+1) subarrays and verify k-1 pairs
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-**Space Complexity:** O(1) excluding the output array, as we use constant extra space
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-**Edge Case:** If k = 1, each single element is a valid subarray with power equal to that element
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**1.2 High-level approach:**
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For each possible subarray of size k, we check if it's sorted in ascending order and contains consecutive integers. If both conditions are met, the power is the maximum (last) element; otherwise, it's -1.
**Restate the problem:** We need to make every element in the array divisible by 3 by adding or subtracting 1. Each operation changes one element by 1, and we want the minimum number of operations.
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**1.1 Constraints & Complexity:**
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- Input size: `1 <= nums.length <= 50`, and each `1 <= nums[i] <= 50`
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-**Time Complexity:** O(n) where n is the length of nums, as we iterate through each element once
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-**Space Complexity:** O(1) as we only use a constant amount of extra space
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-**Edge Case:** If an element is already divisible by 3, no operation is needed
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**1.2 High-level approach:**
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The goal is to count how many elements are not divisible by 3, since each such element requires exactly 1 operation to make it divisible by 3.
-**Brute Force:** For each element, try both adding and subtracting to find the minimum operations. This would be O(n) with unnecessary checks.
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-**Optimized Strategy:** For any number x, if `x % 3 != 0`, we can always make it divisible by 3 with exactly 1 operation (either add `(3 - x % 3)` or subtract `(x % 3)`). This is O(n) with a single pass.
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**1.4 Decomposition:**
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1. Iterate through each number in the array
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2. Check if the number is divisible by 3 using modulo operation
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3. If not divisible, increment the operation count by 1
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4. Return the total count of operations
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### Steps (The "How")
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**2.1 Initialization & Example Setup:**
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Let's use the example: `nums = [1, 2, 3, 4]`
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- Initialize `res = 0` to count operations
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**2.2 Start Checking:**
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We begin checking each element to see if it's divisible by 3.
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**2.3 Trace Walkthrough:**
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Using the example `nums = [1, 2, 3, 4]`:
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| Element | Remainder (num % 3) | Is Divisible? | Operations Needed | res |
**Restate the problem:** We need to count beautiful subarrays. A subarray is beautiful if we can make all its elements equal to 0 using operations that subtract 2^k from pairs of elements that both have the k-th bit set.
-**Time Complexity:** O(n) where n is the length of nums, as we use a single pass with prefix XOR and hash map lookups
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-**Space Complexity:** O(n) in the worst case to store prefix XOR values in the hash map
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-**Edge Case:** A subarray with all zeros is beautiful (XOR = 0)
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**1.2 High-level approach:**
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The key insight is that a subarray is beautiful if and only if its XOR is 0. This is because we can pair up all set bits and subtract them. We use prefix XOR to efficiently compute subarray XORs.
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**1.3 Brute force vs. optimized strategy:**
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-**Brute Force:** Check all O(n²) subarrays and compute XOR for each, which is O(n³) overall.
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-**Optimized Strategy:** Use prefix XOR. If `prefix_xor[i] == prefix_xor[j]`, then the subarray from i to j-1 has XOR 0. This is O(n) with hash map lookups.
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**1.4 Decomposition:**
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1. Initialize prefix XOR to 0 and a hash map to count prefix XOR occurrences
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2. For each element, update prefix XOR by XORing with the current element
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3. Count how many times we've seen this prefix XOR value before (these form beautiful subarrays)
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### Strategy (The "Why")
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**1.1 Constraints & Complexity:**
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**Restate the problem:** We need to count beautiful substrings where vowels equal consonants and their product is divisible by k. A substring is beautiful if it has equal numbers of vowels and consonants, and (vowels × consonants) % k == 0.
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-**Constraints:** $1 \leq n \leq 10^{15}$.
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-**Time Complexity:** $O(d)$ where $d$ is the number of digits in $n$ (at most 16 digits).
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-**Space Complexity:** $O(d)$ for storing the string representation and power array.
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-**Edge Case:** If $n = 1$, return 1. If $n$ contains a zero, we stop counting at that position.
-**Time Complexity:** O(n²) where n is the string length, as we check all possible substrings
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-**Space Complexity:** O(1) as we only use constant extra space for counters
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-**Edge Case:** If k = 1, any substring with vowels == consonants is beautiful since any number is divisible by 1
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**1.2 High-level approach:**
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For each possible starting position, we expand the substring and maintain counts of vowels and consonants. When both counts are equal and their product is divisible by k, we found a beautiful substring.
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The goal is to count distinct integers from $[1, n]$ after removing all zeros from their decimal representation. Key insight: numbers with zeros, when zeros are removed, produce numbers that were already counted. So we need to count only numbers that don't contain any zeros.
-**Brute Force:** Generate all numbers from 1 to $n$, remove zeros, and count distinct values. This is infeasible for $n = 10^{15}$.
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-**Optimized Strategy:** Count numbers without zeros using combinatorics. For each digit position, we can place digits 1-9 (9 choices). Use digit DP or combinatorial counting. This is $O(d)$ time.
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-**Why optimized is better:** The combinatorial approach avoids generating all numbers and works efficiently even for very large $n$.
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-**Brute Force:** Generate all substrings and check each one. This is what we do, O(n²).
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-**Optimized Strategy (for larger constraints):** Use prefix sums and hash maps to count in O(n) time, but for n ≤ 1000, the O(n²) approach is acceptable.
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**1.4 Decomposition:**
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1. Convert $n$ to string for digit-by-digit processing.
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2. Precompute powers of 9 (for digits 1-9, excluding 0).
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3. Count all numbers with fewer digits than $n$: sum of $9^l$ for $l = 1$ to $len(n)-1$.
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4. Count numbers with same number of digits as $n$ but $\leq n$: for each digit position, count valid numbers.
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5. If $n$ itself has no zeros, add 1.
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1. Iterate through all possible starting positions
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2. For each starting position, expand the substring character by character
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3. Maintain counts of vowels and consonants
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4. Check if vowels == consonants and (vowels × consonants) % k == 0
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5. Count valid beautiful substrings
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### Steps (The "How")
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**2.1 Initialization & Example Setup:**
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Let's use the example: `n = 273`
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We convert to string: `s = "273"`, length = 3.
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Let's use the example: `s = "baeyh"`, `k = 2`
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- Define vowels set: `{'a', 'e', 'i', 'o', 'u'}`
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- Initialize `res = 0`
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**2.2 Start Checking:**
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We count numbers with 1 and 2 digits, then count 3-digit numbers $\leq 273$.
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We begin checking substrings starting from index 0.
- Precompute: `pow9[i] = 9^i` for $i = 0$ to $len(s)$
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- Count shorter numbers: `for l in range(1, len(s)): res += pow9[l]`
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- Count same-length numbers: for each digit position $i$:
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- If digit is 0, break (can't form more valid numbers)
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- For each digit $d$ from 1 to $(digit-1)$: `res += pow9[len(s) - i - 1]`
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For each valid beautiful substring found, we increment the result counter. We continue checking all possible substrings.
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**2.5 Return Result:**
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For $n = 273$, we count all numbers from 1 to 273 that don't contain zeros. The result is the number of distinct integers after removing zeros from all numbers in $[1, 273]$.
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After processing all substrings, `res = 2`, which represents the two beautiful substrings: "baey" and "aeyh".
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