You are given an integer array nums containing positive integers. We define a function encrypt such that encrypt(x) replaces every digit in x with the largest digit in x. For example, encrypt(523) = 555 and encrypt(213) = 333.
Return the sum of encrypted elements.
Example 1:
Input: nums = [1,2,3]
Output: 6
Explanation: The encrypted elements are [1,2,3]. The sum of encrypted elements is 1 + 2 + 3 == 6.
Example 2:
Input: nums = [10,21,31]
Output: 66
Explanation: The encrypted elements are [11,22,33]. The sum of encrypted elements is 11 + 22 + 33 == 66.
Constraints:
1 <= nums.length <= 501 <= nums[i] <= 1000This approach involves iterating through the list of numbers, and for each number, finding the largest digit. We replace every digit of the number with this largest digit to form the encrypted number. The sum of these encrypted numbers is then computed.
This C program defines an encrypt function that finds the largest digit of a given number and replaces all digits with this maximum digit. In sumOfEncryptedElements, we iterate over each number, encrypt it, and compute the sum of all encrypted numbers.
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Time Complexity: O(n * m), where n is the number of elements in the array and m is the number of digits in the largest number.
Space Complexity: O(1), as no additional data structures are used outside of function call stack space.
This approach involves precalculating the maximum digit for each number and then constructing the new number using a repetitive process. Though similar to Approach 1, this highlights the preprocessing step distinctly and showcases optimization possibilities in reconstruction.
This solution separates the maximum digit determination and number reconstruction into two distinct functions, allowing flexibility in optimization and readability of the code logic.
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Time Complexity: O(n * m) for both determining the max digit and reconstructing the number. It does not provide an asymptotic improvement over Approach 1.
Space Complexity: O(1).
| Approach | Complexity |
|---|---|
| Approach 1: Iterative Transformation | Time Complexity: O(n * m), where n is the number of elements in the array and m is the number of digits in the largest number. |
| Approach 2: Precalculate Maximum Digit and Rebuild | Time Complexity: O(n * m) for both determining the max digit and reconstructing the number. It does not provide an asymptotic improvement over Approach 1. |
Combination Sum - Backtracking - Leetcode 39 - Python • NeetCode • 372,069 views views
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