Minimum Division Operations to Make Array Non Decreasing

• State Representation: Each unique configuration of the nums array is a state in a graph.
• Edges: An edge exists from state A to state B if B can be reached from A by one operation.
• Goal: Find the shortest path from the initial state to any state where the array is non-decreasing.
• Algorithm: Use Breadth-First Search (BFS) to find this shortest path.
  1. Initialize a queue and add the initial state (initial_nums, 0). A state is a pair of the array and the number of operations.
  2. Use a Set to store visited array configurations to avoid cycles and redundant work.
  3. While the queue is not empty: a. Dequeue (current_array, ops). b. If current_array is non-decreasing, return ops as it's the minimum. c. For each element nums[i] in current_array: i. Check if nums[i] is composite. If so, it can be operated on. ii. Calculate the new value new_val by applying the operation. iii. Create a next_array with this change. iv. If next_array has not been visited, add it to the visited set and enqueue (next_array, ops + 1).

The core idea is to treat every possible configuration of the nums array as a node in a vast graph. An operation that transforms one array configuration into another is a directed edge between the corresponding nodes. Since each operation has a cost of 1, the problem of finding the minimum number of operations is equivalent to finding the shortest path from the initial array configuration to any valid (non-decreasing) configuration. BFS is the standard algorithm for finding the shortest path in an unweighted graph. We start with the initial array and explore all arrays reachable in 1 operation, then all arrays reachable in 2 operations, and so on. A visited set is crucial to prevent re-processing the same array configuration, which would lead to infinite loops and redundant computations. However, the number of possible states is astronomically large, rendering this approach impractical.

• Greedy Choice: Iterate the array from right to left (i from n-2 down to 0). For each i, ensure nums[i] <= nums[i+1].
• Modification: If nums[i] > nums[i+1], we must reduce nums[i].
• Operation: The operation transforms a composite number x into its smallest prime factor (spf(x)). A prime number cannot be changed.
• Algorithm Steps:
  1. Initialize total_operations = 0.
  2. Loop i from n-2 down to 0.
  3. If nums[i] > nums[i+1]: a. Find the smallest prime factor of nums[i], let's call it spf_val, using a helper function that performs trial division (checking divisibility up to sqrt(nums[i])). b. If nums[i] is prime (spf_val == nums[i]), it cannot be reduced. Return -1. c. nums[i] is composite. Apply the operation: nums[i] = spf_val. Increment total_operations. d. The new nums[i] is now prime. If it's still greater than nums[i+1], it's impossible to satisfy the condition. Return -1.
  4. Return total_operations.

This greedy strategy works because decisions made at index i do not invalidate the already-established property for indices greater than i. By processing from right to left, when we consider the pair (nums[i], nums[i+1]), the value of nums[i+1] has already been finalized to be as large as possible while satisfying nums[i+1] <= nums[i+2]. To satisfy nums[i] <= nums[i+1], we must decrease nums[i] if it's larger. Since a composite number becomes prime (and thus unchangeable) after one operation, we have at most one chance to fix nums[i]. We perform the operation and check if the condition is now met. If not, it's impossible. The main bottleneck of this specific implementation is the factorization step.

Here is a code snippet for the helper function and the main logic:

1class Solution {2    private int getSmallestPrimeFactor(int n) {3        if (n <= 1) return n;4        if (n % 2 == 0) return 2;5        for (int i = 3; i * i <= n; i += 2) {6            if (n % i == 0) {7                return i;8            }9        }10        return n; // n is prime11    }12 13    public int minimumOperations(int[] nums) {14        int n = nums.length;15        int operations = 0;16        for (int i = n - 2; i >= 0; i--) {17            if (nums[i] > nums[i+1]) {18                int currentNum = nums[i];19                int spf = getSmallestPrimeFactor(currentNum);20                if (spf == currentNum) { // currentNum is prime21                    return -1;22                }23                currentNum = spf;24                operations++;25                if (currentNum > nums[i+1]) {26                    return -1;27                }28                nums[i] = currentNum;29            }30        }31        return operations;32    }33}

• Pre-computation: Before processing the array, pre-compute the Smallest Prime Factor (SPF) for all numbers up to the maximum possible value (10^6) using a Sieve.
  1. Create an spf array of size 10^6 + 1.
  2. Initialize spf[i] = i.
  3. Iterate p from 2. If spf[p] == p, p is prime. For all multiples j of p, set spf[j] = p if it's not already set.
• Greedy Logic: The core logic is the same as the previous greedy approach, but SPF lookups are now O(1).
  1. Initialize total_operations = 0.
  2. Loop i from n-2 down to 0.
  3. If nums[i] > nums[i+1]: a. Get spf_val = spf[nums[i]] in O(1) time. b. If spf_val == nums[i] (nums[i] is prime), return -1. c. nums[i] becomes spf_val. Increment total_operations. d. If the new nums[i] is still greater than nums[i+1], return -1.
  4. Return total_operations.

The overall strategy remains the same: a right-to-left greedy pass. The key improvement is the pre-computation step. A sieve algorithm can efficiently find the smallest prime factor for every number up to a limit M. This is done once. After this, the main logic can proceed, and whenever it needs the smallest prime factor of a number x, it can simply look it up in the pre-computed spf array at index x. This reduces the complexity of each step inside the main loop from O(sqrt(M)) to O(1), leading to a much better overall time complexity.

Here is a sample implementation:

1class Solution {2    private static final int MAX_VAL = 1000001;3    private static final int[] spf = new int[MAX_VAL];4 5    // Static initializer block to run the sieve once per class load6    static {7        for (int i = 0; i < MAX_VAL; i++) {8            spf[i] = i;9        }10        for (int i = 2; i * i < MAX_VAL; i++) {11            if (spf[i] == i) { // i is a prime number12                for (int j = i * i; j < MAX_VAL; j += i) {13                    if (spf[j] == j) { // if spf[j] is not set yet14                        spf[j] = i;15                    }16                }17            }18        }19    }20 21    public int minimumOperations(int[] nums) {22        int n = nums.length;23        int operations = 0;24        // We need to create a copy if we cannot modify the input, 25        // but problem context implies we can. Let's assume modification is allowed.26        for (int i = n - 2; i >= 0; i--) {27            if (nums[i] > nums[i+1]) {28                int currentNum = nums[i];29                // If currentNum is prime (and > 1), it cannot be reduced.30                if (spf[currentNum] == currentNum) { 31                    return -1;32                }33                // Perform one operation: num becomes its smallest prime factor.34                currentNum = spf[currentNum];35                operations++;36                // After one op, the number is prime. If it's still too large, impossible.37                if (currentNum > nums[i+1]) {38                    return -1;39                }40                nums[i] = currentNum;41            }42        }43        return operations;44    }45}