1558. Minimum Numbers of Function Calls to Make Target Array

MediumGreedyBit ManipulationArray

Problem Description

You are provided with two integer arrays. The first one is called nums, and it contains some integers. The second one is called arr, which is the same length as nums but initially filled with 0s. Your goal is to make arr look exactly like nums.

To achieve your goal, you are allowed to use a special modify function, which can perform one of two operations:

Increment by 1: Choose any index i and increase the value of arr[i] by 1.
Double all: Double the value of each element in the array arr.

Your task is to figure out the minimum number of calls to the modify function required to transform arr into nums. The end goal is to achieve this transformation with the smallest possible number of operations.

Intuition

The solution strategy revolves around working backwards from nums to arr, as decreasing or halving elements is easier to track than incrementing or doubling. The insight is that any number in nums is formed by starting with 0, potentially doubling several times, and then incrementing.

To minimize calls to modify, we should maximize the use of the doubling operation, because each doubling operation is equivalent to many increment operations. Thus, the strategy is to:

Find the total number of increment operations needed for each number in nums by counting the number of set bits (since each set bit represents an increment operation that has occurred).
Find the maximum number of doubling operations needed. This is achieved by determining the highest power of 2 required to reach the maximum number in nums, which is equivalent to the bit length of the maximum number minus 1 (since starting from 0 and doubling does not count as an operation).

By combining the count of set bits across all numbers with the maximum bit length, we arrive at the minimum number of operations needed to transform arr into nums.

Learn more about Greedy patterns.

Solution Approach

The solution uses simple bitwise operations to find the answer. Here's how the implementation goes by breaking it down into logical steps:

Iterate through each number in the given nums array to calculate the total number of increment operations for each number and the highest number of doubling operations for the whole array.
For each number, we use the bitwise operation bit_count(), which returns the count of set bits (1-bits) in the number. The count of set bits actually represents how many times the increment operation has been performed to reach this number from 0 because each increment operation can only set a bit from 0 to 1.
To get the maximum number of doubling operations across all numbers in the array, we look for the maximum value in nums. We call bit_length() on this number, which gives us the position of the highest set bit, i.e., the number of times we can potentially divide the number by 2 until it reaches 0. This represents the doubling operations needed to reach this number from 1. Since we start with 0, not 1, we subtract 1 from the bit length as the initial state does not count as a doubling operation.
The total minimum number of function calls is the sum of all increment operations (sum(v.bit_count() for v in nums)) and the maximum number of doubling operations (max(0, max(nums).bit_length() - 1)). We also ensure that the number of doubling operations is not negative (which can happen if nums contains all zeros), by taking max(0, ...).

The Python implementation provided uses a list comprehension to calculate the total count of bits for all numbers in nums and finds the maximum bit length among all numbers. Then it adds these two values to get the final answer.

Here is what the code does, broken down step by step:

sum(v.bit_count() for v in nums) - This part goes through every number in nums and accumulates the total increment operations required by summing up the number of set bits in each of the numbers.
max(nums).bit_length() - This part finds the maximum number in nums and then calculates the length of the bits needed to represent that number, which corresponds to the number of doubling operations needed plus one (since index starts at zero).
Subtracting 1 from the bit length - We subtract 1 because the sequence starts from 0 and doubling from 0 does not change the number, so the first doubling operation effectively starts when going from 1 to 2.

Therefore, by combining these steps, the implementation efficiently computes the minimum number of operations with a time complexity of O(n) where n is the length of the nums array.

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Example Walkthrough

Let's assume we have the following input:

nums array: [3, 8, 1]

We want to convert the arr array: [0, 0, 0] to look exactly like nums.

Now, let's walk through the solution approach for this example:

Increment Operations:
- For the number 3, it has a binary representation of 11, which has 2 set bits.
- For the number 8, it has a binary representation of 1000, which has 1 set bit.
- For the number 1, it has a binary representation of 1, which has 1 set bit.
So the total number of increment operations needed is 2 (for 3) + 1 (for 8) + 1 (for 1) = 4.
Doubling Operations:
- The maximum number in nums is 8, which has a binary representation of 1000. The bit length is 4. Therefore, we would need 3 doubling operations to get from 1 to 8. Since we start from 0, we subtract one, leaving us with 4 - 1 = 3 doubling operations.

Now, combining both the increment and doubling operations:

Total increment operations = 4 (from step 1)
Total doubling operations = 3 (from step 2)

Therefore, the minimum number of function calls required to transform arr into nums is 4 + 3 = 7.

This walk through shows that we need 7 modify operations to make arr look exactly like nums, which are [3, 8, 1], by incrementing and doubling appropriately.

Solution Implementation

1class Solution:
2    def minOperations(self, nums: List[int]) -> int:
3        # Calculate the total number of set bits (1s) in all the numbers.
4        # This corresponds to the number of increment operations needed.
5        num_of_increment_ops = sum(bin(num).count('1') for num in nums)
6
7        # Find the maximum number in the list to determine the number of
8        # double operations needed. 
9        # We need to find the position of the highest set bit (most significant bit),
10        # which determines the number of times we need to double from 1 to reach 
11        # that value. This is equivalent to the bit length of the maximum number minus 1
12        # since starting from 1 (2^0) requires no doubling operations.
13        highest_num = max(nums) if nums else 0
14        num_of_double_ops = highest_num.bit_length() - 1 if highest_num > 0 else 0
15
16        # The total number of operations is the sum of the increment and double operations.
17        return num_of_increment_ops + num_of_double_ops
18

1class Solution {
2
3    /**
4     * Method to calculate minimum operations to make all elements of nums equal to zero.
5     * An operation is defined as either:
6     *   1) Subtract 1 from an element if it's not 0 or,
7     *   2) Divide all elements by 2 if all elements are even
8     *
9     * @param nums array of integers
10     * @return the minimum number of operations required
11     */
12    public int minOperations(int[] nums) {
13        int totalOperations = 0; // to track the total minimum operations
14        int maxNumber = 0; // to track the maximum number in the array
15
16        // Calculate the sum of bit counts (number of 1's) for all numbers
17        // and find the maximum number simultaneously
18        for (int number : nums) {
19            maxNumber = Math.max(maxNumber, number);
20            totalOperations += Integer.bitCount(number); // add number of 1's in binary representation
21        }
22
23        // Add the length of the binary string of the maximum number minus 1
24        // which accounts for the division by 2 operations
25        totalOperations += Integer.toBinaryString(maxNumber).length() - 1;
26
27        return totalOperations;
28    }
29}
30

1#include <vector>
2#include <algorithm> // For std::max
3
4class Solution {
5public:
6    int minOperations(std::vector<int>& nums) {
7        int operationsCount = 0; // Initialize counter for minimum operations
8        int maxNum = 0; // Variable to store the maximum value in nums
9
10        // Loop through all numbers in the vector
11        for (int value : nums) {
12            // Update maxNum to the maximum value encountered so far
13            maxNum = std::max(maxNum, value);
14            // Add the number of set bits (1s) in the current value to operationsCount 
15            operationsCount += __builtin_popcount(value);
16        }
17
18        // If maxNum is greater than 0, add the number of bits to reach the most significant bit
19        // of the largest number in nums (the number of left shifts needed for the largest number)
20        if (maxNum) {
21            operationsCount += 31 - __builtin_clz(maxNum);
22        }
23
24        // Return the total count of operations needed
25        return operationsCount;
26    }
27};
28

1// Importing Array object 
2import { max } from 'lodash';
3
4let operationsCount: number = 0; // Initialize counter for minimum operations
5let maxNum: number = 0; // Variable to store the maximum value in nums
6
7/**
8 * Calculate the minimum number of operations to make all element of nums equal to zero.
9 * @param {number[]} nums - The array of numbers to be processed.
10 * @return {number} The minimum number of operations required.
11 */
12function minOperations(nums: number[]): number {
13    operationsCount = 0; // Reset the counter for each function call
14    maxNum = 0; 
15
16    // Loop through all numbers in the array
17    for (let value of nums) {
18        // Update maxNum to the maximum value encountered so far
19        maxNum = max([maxNum, value]) || 0;
20        // Add the number of set bits (1s) in the current value to operationsCount 
21        operationsCount += countSetBits(value);
22    }
23
24    // If maxNum is greater than 0, add the number of bits to reach the most significant bit
25    // (the number of left shifts needed for the largest number)
26    if (maxNum) {
27        operationsCount += 31 - clz(maxNum);
28    }
29
30    // Return the total count of operations needed
31    return operationsCount;
32}
33
34/**
35 * Count the number of set bits in an integer (Naive method).
36 * @param {number} n - The number in which set bits are to be counted.
37 * @return {number} The count of set bits in n.
38 */
39function countSetBits(n: number): number {
40    let count: number = 0;
41    while (n) {
42        count += n & 1; // Increment count if the least significant bit is set
43        n = n >>> 1; // Right shift n by 1 bit to process the next bit
44    }
45    return count;
46}
47
48/**
49 * Compute the number of leading zero bits in the integer’s binary representation.
50 * @param {number} num - The number whose leading zeros are to be counted.
51 * @return {number} The count of leading zero bits in num.
52 */
53function clz(num: number): number {
54    if (num === 0) return 32;
55    let leadingZeros = 0;
56    for (let i = 31; i >= 0; i--) {
57        if ((num & (1 << i)) === 0) {
58            leadingZeros++;
59        } else {
60            break;
61        }
62    }
63    return leadingZeros;
64}
65

Time and Space Complexity

The time complexity of the code can be calculated based on two operations: the calculation of bit counts for all numbers and finding the maximum number to calculate its bit length. Calculating the bit count for a single number is O(1), and we do this for every number in the list, resulting in O(n) complexity where n is the length of the list. Finding the maximum value in the list also takes O(n) time. The bit_length function is also O(1) as it typically involves calculating the position of the highest bit set in the integer representation, which does not depend on the size of the number itself but on the number of bits.

Thus, the time complexity of the function is O(n) where n is the size of the nums list.

As for the space complexity, since no additional significant space is used apart from a few variables to store intermediate results, the space complexity is O(1).

In summary:

Time Complexity: O(n)
Space Complexity: O(1)

Learn more about how to find time and space complexity quickly using problem constraints.

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