Minimum Number of Operations to Make All Array Elements Equal to 1

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from math import gcd

class Solution:
    def minOperations(self, nums):
        n = len(nums)
        c = nums.count(1)  # if there is at least 1, then it's easy!
        if c != 0:
            return n - c
        res = 1e7  # just a dummy value
        for i in range(n):
            g = nums[i]
            for j in range(i + 1, n):
                g = gcd(g, nums[j])
                if g == 1:
                    res = min(res, j - i + (n - 1)) # number of operations to make this element 1 + number of non-ones (i.e. n-1).
                    break
        return res if res != 1e7 else -1

10 Prerequisite LeetCode Problems

For this, the following are a good preparation:

  1. “326. Power of Three” - Understanding how to factorize numbers, which is needed to calculate the gcd.

  2. “204. Count Primes” - Understanding the basics of prime numbers, which is important when finding gcd.

  3. “50. Pow(x, n)” - Understanding the concept of exponentiation, which can be useful when thinking about mathematical operations.

  4. “231. Power of Two” - Another problem about powers that can help with understanding how to break down numbers into their factors.

  5. “1015. Smallest Integer Divisible by K” - Understanding the concept of divisibility, which is critical in the process of finding gcd.

  6. “365. Water and Jug Problem” - It has a similar mathematical problem of finding gcd.

  7. “914. X of a Kind in a Deck of Cards” - Practices finding the gcd in a set of numbers.

  8. “1071. Greatest Common Divisor of Strings” - This problem involves finding the gcd, but with strings instead of numbers.

  9. “400. Nth Digit” - The problem involves dealing with number sequences and finding patterns, a skill that is helpful for this problem.

  10. “441. Arranging Coins” - This problem involves a kind of operation counter, similar to the main problem’s requirement of counting the number of operations.

These cover number theory such as prime numbers, greatest common divisors, and factorization.

You are given a 0-indexed array nums consisiting of positive integers. You can do the following operation on the array any number of times:

Select an index i such that 0 <= i < n - 1 and replace either of nums[i] or nums[i+1] with their gcd value. Return the minimum number of operations to make all elements of nums equal to 1. If it is impossible, return -1.

The gcd of two integers is the greatest common divisor of the two integers.

Example 1:

Input: nums = [2,6,3,4] Output: 4 Explanation: We can do the following operations:

  • Choose index i = 2 and replace nums[2] with gcd(3,4) = 1. Now we have nums = [2,6,1,4].
  • Choose index i = 1 and replace nums[1] with gcd(6,1) = 1. Now we have nums = [2,1,1,4].
  • Choose index i = 0 and replace nums[0] with gcd(2,1) = 1. Now we have nums = [1,1,1,4].
  • Choose index i = 2 and replace nums[3] with gcd(1,4) = 1. Now we have nums = [1,1,1,1]. Example 2:

Input: nums = [2,10,6,14] Output: -1 Explanation: It can be shown that it is impossible to make all the elements equal to 1.

Constraints:

2 <= nums.length <= 50 1 <= nums[i] <= 106

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from math import gcd
from typing import List

class Solution:
    def minOperations(self, nums: List[int]) -> int:
        n = len(nums)
        c = nums.count(1)     # if there is at least 1. then it's easy!
        if c != 0:    
            return n - c
        res = 1e7        # just a dummy value
        for i in range(n):
            g = nums[i]
            for j in range(i+1, n):
                g = gcd(g, nums[j])
                if g == 1:
                    res = min(res, j - i + (n - 1)) # number of operations to make this element 1+ number of non ones (i.e. n-1) .
                    break
        return -1 if res == 1e7 else res

  1. We can make all elements equal to 1 by replacing a non-1 adjacent number with 1. Hence, the problem reduces to finding a subarray (contiguous segment) in the array such that the gcd of the numbers in the subarray is 1.

  2. To solve this problem efficiently, we need to iterate through all possible subarrays of the array. This requires a nested loop, resulting in a time complexity of O(n^2), where n is the size of the array.

  3. For each subarray, we calculate the gcd of the elements in the subarray. If the gcd is 1, we can make all the elements of the original array equal to 1 in n - size_of_subarray operations. The goal is to minimize the number of operations, so we track the minimum size of such a subarray.

  4. If we never find a subarray with a gcd of 1, we return -1 as it is impossible to make all elements equal to 1.

This approach efficiently solves the problem by exploiting the properties of the gcd operation and the characteristics of the problem.

Problem Classification

The problem is a computational problem that involves array manipulation and concepts from number theory, namely the calculation of the greatest common divisor (gcd).

‘What’ components:

  • An input, which is a 0-indexed array of positive integers, nums.
  • A series of operations that can be performed on this array, where for each operation, an index i is selected (0 <= i < n - 1) and either nums[i] or nums[i+1] is replaced with their gcd.
  • The desired output, which is the minimum number of operations needed to make all elements of nums equal to 1. If this is impossible, the function should return -1.

Categorization:

  1. Domain-Based Categorization: This problem statement falls under the domain of Number Theory and Array Manipulation in Computer Science.

  2. Based on ‘What’ components:

    • Array: The problem involves manipulating an array of integers.
    • Number Theory: The problem involves calculating the gcd of two integers, which is a concept in number theory.
    • Optimization: The problem requires finding the minimum number of operations, which is a type of optimization problem.

Thus, this problem can be classified as an Optimization problem in Array Manipulation and Number Theory.

Language Agnostic Coding Drills

  1. Identify the distinct coding concepts in the given code:

    • Variable initialization: Assigning initial values to variables.
    • List operations: Accessing elements in a list, using list functions like count.
    • Use of if-else conditions: Implementing conditional branches in code.
    • Use of loops: Writing for loops to iterate over sequences.
    • Use of nested loops: Using one loop inside another to handle complex cases.
    • Math operations: Using mathematical operations like min, gcd.
    • Function definition: Defining a function with input parameters and a return value.
    • Use of Python specific syntax: from for importing specific functions from a module, type hints (List[int], int), etc.
    • Early exit from loops: Using break to terminate a loop prematurely.

  2. Arrange these coding drills in increasing order of difficulty:

    • Variable initialization: Basic concept to hold and store values.
    • Use of if-else conditions: Requires understanding of conditional logic.
    • List operations: Need knowledge of data structures.
    • Math operations: Depends on understanding of mathematical concepts.
    • Function definition: Involves understanding how to encapsulate logic within a function.
    • Use of loops: Involves control flow and potentially more complex logic.
    • Use of Python specific syntax: Understanding Python syntax and semantics.
    • Use of nested loops: Complexity increases with nested loops due to higher time complexity.
    • Early exit from loops: This involves understanding when and how to exit a loop prematurely based on certain conditions.

  3. Problem-solving approach:

    • Read and understand the problem statement. Identify the input, the required output, and any constraints.
    • Initialize variables. In this case, the variable n is set to the length of the input list, c counts the number of 1s in the list, and res is set to a high dummy value.
    • Implement the simple case first. If c is not 0, return n - c.
    • For the more complex case, iterate over the list with a for loop. This serves as a drill to practice using loops.
    • Within this loop, iterate over the rest of the list from i+1 to n with another for loop, forming a nested loop to practice handling such scenarios.
    • Use the gcd function within the inner loop to find the greatest common divisor of g and nums[j]. This is a math operation that is useful in many algorithms.
    • Implement a conditional check to see if g is 1. If it is, update res to be the minimum of its current value and j - i + (n - 1), then break the inner loop. This involves the use of if-else conditions, the min function, and early exit from a loop.
    • Finally, after the loops, return -1 if res is still the dummy value, otherwise return res. This is the final result of the function.

Each drill is necessary and contributes to the overall solution. They provide the means to traverse the list, perform the necessary calculations, and return the correct result.

Targeted Drills in Python

  1. Separate Python drills for each identified concept:

    • Variable initialization:
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    x = 5
y = 'Hello World'
    • Use of if-else conditions:
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    x = 5
if x > 0:
    print('Positive')
else:
    print('Non-Positive')
    • List operations:
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    list1 = [1, 2, 3, 1]
print(list1.count(1))  # prints 2
    • Math operations:
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    from math import gcd, min
print(gcd(6, 9))  # prints 3
print(min(3, 5))  # prints 3
    • Function definition:
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    def say_hello(name):
    return f'Hello {name}!'
    • Use of loops:
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    for i in range(5):
    print(i)
    • Use of nested loops:
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    for i in range(5):
    for j in range(5):
        print(i, j)
    • Early exit from loops:
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    for i in range(5):
    if i == 3:
        break
    print(i)  # prints 0, 1, 2

  2. Problem-specific concepts and their drills:

    • The problem-specific concept in this case is using gcd function inside nested loops and updating a variable res based on conditions.
    • Here’s a drill:
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    nums = [2, 6, 3, 4]
res = 1e7
for i in range(len(nums)):
    g = nums[i]
    for j in range(i+1, len(nums)):
        g = gcd(g, nums[j])
        if g == 1:
            res = min(res, j - i + 1)
            break
print(-1 if res == 1e7 else res)  # prints 4

    This drill is essential as it captures the core logic of our problem - traversing a list, calculating gcd in a nested loop, and updating a result variable based on conditions.

  3. Integrating the drills:

    • Start by importing necessary functions and defining the function with the appropriate parameters (Function definition drill).
    • Initialize variables n, c and res (Variable initialization drill).
    • Add an if-else condition to return n - c if c is not 0 (If-else condition drill).
    • Next, implement the nested loop structure (Nested loops drill).
    • Inside the inner loop, compute gcd and check if it’s 1 (Math operations and If-else condition drills).
    • If it’s 1, update res and break the loop (Early exit from loops drill).
    • Finally, after the loops, add a return statement based on the value of res (If-else condition drill).

The final solution would integrate all these drills in the order mentioned above. Each drill plays a vital role in building up to the solution.