Apply Operations to Make Sum of Array Greater Than or Equal to k

• Handle the edge case: if k is 1, the initial array [1] is sufficient, so return 0.
• The total number of operations is the sum of increment operations and duplicate operations. A known upper bound for the minimum operations is k - 1 (achieved by either incrementing the initial 1 to k, or by duplicating the 1 k-1 times).
• We can iterate through all possible values v that the elements in the final array could have. A reasonable range for v is from 1 to k.
• For each value v, we determine the minimum number of elements c needed to make the sum at least k. This is given by c = ceil(k / v), which can be calculated using integer arithmetic as c = (k + v - 1) / v.
• The number of increment operations to get value v is v - 1.
• The number of duplicate operations to get c elements is c - 1.
• The total operations for a given v is (v - 1) + (c - 1).
• We keep track of the minimum total operations found across all tested values of v.
• After checking all v from 1 to k, the minimum value found is the answer.

The fundamental insight is that the most efficient way to construct the final array is to have all its elements be identical. Let's say we decide to create an array of c elements, each with a value of v.

To achieve this from the initial array [1], we need to perform two types of operations:

1. Increment Operations: To change the initial element 1 to v, we need v - 1 increments.
2. Duplicate Operations: To get an array of c elements from a single element, we need c - 1 duplications.

The total number of operations is (v - 1) + (c - 1).

The sum of the array elements will be v * c. To satisfy the problem condition, we must have v * c >= k.

To minimize operations for a fixed v, we should choose the smallest possible integer c that satisfies the condition. This value is c = ceil(k / v). In integer arithmetic, this is calculated as c = (k + v - 1) / v.

This approach iterates through all possible values of v from 1 up to k, calculates the corresponding c and the total operations, and finds the minimum among them.

1class Solution {2    public int minOperations(int k) {3        if (k == 1) {4            return 0;5        }6        // An initial upper bound is k-1 operations.7        // This can be achieved by setting v=k, c=1 (k-1 increments, 0 duplicates)8        // or v=1, c=k (0 increments, k-1 duplicates).9        int minOps = k - 1;10 11        // Iterate through all possible values 'v' from 1 to k.12        for (int v = 1; v <= k; v++) {13            // For a given v, the minimum count 'c' needed is ceil(k/v).14            int c = (k + v - 1) / v;15            16            // Total operations = (increment ops) + (duplicate ops)17            int currentOps = (v - 1) + (c - 1);18            19            minOps = Math.min(minOps, currentOps);20        }21        22        return minOps;23    }24}

• The problem is to minimize (v - 1) + (c - 1) subject to v * c >= k.
• The optimal integer solution (v, c) for this problem will have the property that min(v, c) <= sqrt(k).
• This insight allows us to avoid checking all v up to k. We can split the search into two parts:
  1. Cases where v <= sqrt(k).
  2. Cases where c <= sqrt(k) (which covers the remaining cases where v > sqrt(k)).
• First, iterate v from 1 up to floor(sqrt(k)). For each v, calculate c = ceil(k/v) and the total operations, updating the minimum.
• Second, iterate c from 1 up to floor(sqrt(k)). For each c, calculate the required v = ceil(k/c) and the total operations, updating the minimum.
• The overall minimum found after these two loops is the answer.

We are trying to minimize the function ops(v, c) = v + c - 2 subject to v * c >= k.

The expression v + c is minimized when v and c are close to each other. For a fixed product v*c = P, the sum v+c is minimized when v = c = sqrt(P). This suggests that our optimal integer solution (v, c) will be close to the point (sqrt(k), sqrt(k)) on the v-c plane.

Let's prove that for an optimal solution (v_opt, c_opt), we must have min(v_opt, c_opt) <= sqrt(k). Assume for contradiction that v_opt > sqrt(k) and c_opt > sqrt(k). Then their product v_opt * c_opt > k. This means the solution is not 'tight'. We could potentially decrease v_opt to v_opt - 1 and still satisfy the sum constraint, while reducing the total operations. A more formal proof shows that one of the two must be less than or equal to sqrt(k).

This observation allows us to limit our search. We only need to check pairs (v, c) where at least one component is small.

1. We iterate through all small values of v (i.e., v from 1 to sqrt(k)). For each v, we find the best c and calculate the operations.
2. We iterate through all small values of c (i.e., c from 1 to sqrt(k)). For each c, we find the best v and calculate the operations. This second loop covers the cases where v is large (since if v > sqrt(k), then c must be small).

1class Solution {2    public int minOperations(int k) {3        if (k == 1) {4            return 0;5        }6        int minOps = k - 1;7        int limit = (int) Math.sqrt(k);8 9        // Case 1: Iterate through v from 1 to sqrt(k).10        // This covers all pairs where v is small.11        for (int v = 1; v <= limit; v++) {12            int c = (k + v - 1) / v;13            minOps = Math.min(minOps, (v - 1) + (c - 1));14        }15 16        // Case 2: Iterate through c from 1 to sqrt(k).17        // This covers all pairs where c is small (and v is large).18        // Note: There is some overlap with Case 1, but it's harmless.19        for (int c = 1; c <= limit; c++) {20            int v = (k + c - 1) / c;21            minOps = Math.min(minOps, (v - 1) + (c - 1));22        }23 24        return minOps;25    }26}

• The function to minimize is f(v) = (v - 1) + (ceil(k / v) - 1).
• This function is unimodal, meaning it has a single minimum value over the search range [1, k].
• We can use ternary search to find the value of v that minimizes this function.
• Initialize a search range low = 1 and high = k.
• Repeatedly shrink the search range:
  • Calculate two midpoints m1 = low + (high - low) / 3 and m2 = high - (high - low) / 3.
  • Compare f(m1) and f(m2).
  • If f(m1) < f(m2), the minimum must lie in the interval [low, m2], so we set high = m2.
  • Otherwise, the minimum must lie in [m1, high], so we set low = m1.
• Continue this until the range [low, high] is very small (e.g., high - low < 3).
• Finally, perform a linear scan over the small remaining interval [low, high] to find the exact minimum.

The cost function we want to minimize is f(v) = (v - 1) + (ceil(k / v) - 1). Let's analyze its components:

• v - 1: This term increases linearly with v.
• ceil(k / v) - 1: This term decreases as v increases.

The sum of a strictly increasing function and a strictly decreasing function results in a convex or unimodal function. The ceil operation introduces some steps, but the overall shape remains unimodal. This means there's a single valley, and we can use an efficient search algorithm to find its bottom.

Ternary search is designed for this exact scenario. It's similar to binary search but divides the search space into three parts instead of two. By evaluating the function at two points (m1 and m2), it can eliminate one-third of the search space in each step, leading to a logarithmic time complexity.

1class Solution {2    public int minOperations(int k) {3        if (k == 1) {4            return 0;5        }6 7        long low = 1, high = k;8        long minOps = k - 1; // Initial upper bound9 10        // Ternary search to find the 'v' that minimizes the operations.11        // The function f(v) = (v-1) + (ceil(k/v)-1) is unimodal.12        while (high - low >= 3) {13            long m1 = low + (high - low) / 3;14            long m2 = high - (high - low) / 3;15            if (calculateOps(m1, k) < calculateOps(m2, k)) {16                high = m2;17            } else {18                low = m1;19            }20        }21 22        // After the loop, the minimum is in the small range [low, high].23        // We check this range exhaustively.24        for (long v = low; v <= high; v++) {25            minOps = Math.min(minOps, calculateOps(v, k));26        }27 28        return (int) minOps;29    }30 31    // Helper function to calculate operations for a given value 'v'.32    private long calculateOps(long v, int k) {33        if (v == 0) return Long.MAX_VALUE;34        long c = (k + v - 1) / v; // c = ceil(k/v)35        return (v - 1) + (c - 1);36    }37}