Build Array Where You Can Find The Maximum Exactly K Comparisons

class Solution {
public:
    int countWays(int n, int m, int k) {
        long long current[m + 1][k + 1];
        long long currentSum[m + 1][k + 1];
        long long previous[m + 1][k + 1];
        long long prevSum[m + 1][k + 1];
        memset(current, 0, sizeof(current));
        memset(currentSum, 0, sizeof(currentSum));
        memset(previous, 0, sizeof(previous));
        memset(prevSum, 0, sizeof(prevSum));
        int MODULO = 1000000007;
            
        for (int number = 1; number <= m; number++) {
            current[number][1] = 1;
        }
            
        for (int idx = 1; idx <= n; idx++) {
            if (idx > 1) {
                memset(current, 0, sizeof(current));
            }
                
            memset(currentSum, 0, sizeof(currentSum));
                
            for (int maxValue = 1; maxValue <= m; maxValue++) {
                for (int complexity = 1; complexity <= k; complexity++) {
                    long long value = (maxValue * previous[maxValue][complexity]) % MODULO;
                    value = (value + prevSum[maxValue - 1][complexity - 1]) % MODULO;
    
                    current[maxValue][complexity] += value;
                    current[maxValue][complexity] %= MODULO;
                        
                    currentSum[maxValue][complexity] = (currentSum[maxValue - 1][complexity] + current[maxValue][complexity]);
                    currentSum[maxValue][complexity] %= MODULO;
                }
            }
                
            for (int maxValue = 0; maxValue <= m; maxValue++) {
                for (int complexity = 0; complexity <= k; complexity++) {
                    previous[maxValue][complexity] = current[maxValue][complexity];
                    prevSum[maxValue][complexity] = currentSum[maxValue][complexity];
                }
            }
        }
            
        return currentSum[m][k];
    }
};

The provided code in C++ is a solution to the problem of building an array where you can find the maximum element exactly k times using dynamic programming. Here is a concise explanation of how the code functions:

• Start by defining four 2D arrays: current, currentSum, previous, and prevSum, and initialize them to zero. The current and currentSum arrays store temporal values during computation before transferring them to previous and prevSum, respectively.

• Establish a MODULO constant set to 1000000007 (a large prime number used to avoid integer overflow).

• Initialize the base case for when the number of comparisons (k) is one. If k equals one, any element by itself will reach this condition.

• Nested looping structure:

  • The outermost loop iterates over each position idx from 1 to n.
  • The next loop iterates through possible maximum values maxValue from 1 to m.
  • The innermost loop iterates over the complexity (or number of comparisons) complexity from 1 to k.

• Inside these loops, calculate ways to build arrays considering both constraints (idx and k) by using previously computed values from previous and prevSum.

• Update values in current and currentSum arrays based on these calculations, applying modular arithmetic to ensure values do not exceed the limits of regular integer types.

• After each idx iteration, copy the values from current and currentSum arrays into previous and prevSum arrays respectively to use them in the next iteration over idx.

• Finally, the function returns the value in currentSum[m][k], which represents the total ways to construct the array according to the given conditions.

Ensure you understand the use of both dynamic array states (current/previous) to avoid modifications influencing current iteration results and the importance of modular arithmetic in managing large outputs.

class Calculator {
    public int countSolutions(int length, int maxVal, int complexity) {
        long[][] dynamicTable = new long[maxVal + 1][complexity + 1];
        long[][] sumTable = new long[maxVal + 1][complexity + 1];
        long[][] oldDynamic = new long[maxVal + 1][complexity + 1];
        long[][] oldSum = new long[maxVal + 1][complexity + 1];
        int MODULO = (int) 1e9 + 7;
            
        for (int value = 1; value <= maxVal; value++) {
            dynamicTable[value][1] = 1;
        }
            
        for (int idx = 1; idx <= length; idx++) {
            if (idx > 1) {
                dynamicTable = new long[maxVal + 1][complexity + 1];
            }
                
            sumTable = new long[maxVal + 1][complexity + 1];
                
            for (int maximum = 1; maximum <= maxVal; maximum++) {
                for (int cost = 1; cost <= complexity; cost++) {
                    long result = (maximum * oldDynamic[maximum][cost]) % MODULO;
                    result = (result + oldSum[maximum - 1][cost - 1]) % MODULO;
    
                    dynamicTable[maximum][cost] += result;
                    dynamicTable[maximum][cost] %= MODULO;
                        
                    sumTable[maximum][cost] = (sumTable[maximum - 1][cost] + dynamicTable[maximum][cost]);
                    sumTable[maximum][cost] %= MODULO;
                }
            }
                
            oldDynamic = dynamicTable;
            oldSum = sumTable;
        }
            
        return (int) sumTable[maxVal][complexity];
    }
}

This Java solution tackles the problem of building an array where you can find the maximum value exactly k comparisons by using dynamic programming to keep track of count distributions.

• The function countSolutions is defined, taking three parameters: length, maxVal, and complexity. These correspond to the length of the target array, the maximum value each element in the array can take, and the desired number of maximum comparisons, respectively.
• The solution utilizes four 2D arrays: dynamicTable, sumTable, oldDynamic, and oldSum to preserve states and transitional sums across different iterations.
• The MODULO constant ensures that all numerical operations adhere to modulo (10^9 + 7) to avoid integer overflow.
• Initialization begins by setting every value in dynamicTable[value][1] to 1, assuming that each value is reachable with exactly 1 comparison for the starting position.
• Iteration over the length of the array follows. At each step, new tables for dynamicTable and sumTable are initialized to guarantee freshness of data.
• The nested loops over maximum and cost update the dynamicTable using the formula that combines the current element's contribution and accumulative results from smaller elements and lower costs realized in oldSum.
• Each result is computed modulo MODULO to maintain manageable numbers.
• After processing, the utility switches the current tables (dynamicTable and sumTable) to old ones (oldDynamic and oldSum) for use as reference in the next iteration.
• Finally, the number at sumTable[maxVal][complexity] is returned, representing the total number of arrays of length length where the maximum value occurs with complexity comparisons.

This approach is time-efficient and allows reuse of results due to its dynamic programming nature, partitioning complex problems into simpler, overlapping subproblems managed through tabulation.

class ArrayBuilder:
    def countConfigurations(self, length: int, max_val: int, cost: int) -> int:
        dp_table = [[0] * (cost + 1) for _ in range(max_val + 1)]
        prefix_sum = [[0] * (cost + 1) for _ in range(max_val + 1)]
        previous_dp = [[0] * (cost + 1) for _ in range(max_val + 1)]
        previous_prefix = [[0] * (cost + 1) for _ in range(max_val + 1)]
        MOD_CONST = 10 ** 9 + 7
            
        for value in range(1, max_val + 1):
            dp_table[value][1] = 1
    
        for iteration in range(1, length + 1):
            if iteration > 1:
                dp_table = [[0] * (cost + 1) for _ in range(max_val + 1)]
    
            prefix_sum = [[0] * (cost + 1) for _ in range(max_val + 1)]
            for highest_value in range(1, max_val + 1):
                for curr_cost in range(1, cost + 1):                    
                    computation = (highest_value * previous_dp[highest_value][curr_cost]) % MOD_CONST
                    computation = (computation + previous_prefix[highest_value - 1][curr_cost - 1]) % MOD_CONST
    
                    dp_table[highest_value][curr_cost] += computation
                    dp_table[highest_value][curr_cost] %= MOD_CONST
                        
                    prefix_sum[highest_value][curr_cost] = (prefix_sum[highest_value - 1][curr_cost] + dp_table[highest_value][curr_cost]) % MOD_CONST
                        
            previous_dp = dp_table
            previous_prefix = prefix_sum
    
        return prefix_sum[max_val][cost]

This Python solution outlines how to find the number of configurations to build an array where the maximum value achieves a specified number of comparisons exactly k times. The solution uses dynamic programming with a complexity optimized using a prefix sum technique. Below is a detailed breakdown of the implementation steps and logic.

1. Define the countConfigurations method with parameters: length, max_val, and cost, representing the length of the array, the range's maximum value, and k comparisons, respectively.
2. Initialize the dynamic programming table dp_table, another table prefix_sum to store cumulative values, and two temporary storage lists previous_dp and previous_prefix. All these tables are initialized to zero and have the dimensions based on max_val and cost.
3. Set up a modulus constant, MOD_CONST, to handle large number calculations within a manageable integer range.
4. The base case initialization: Set all configurations of length 1 to have exactly one way to achieve any count up to the cost with each possible maximum value.
5. Loop through possible array lengths. Within each iteration, reset the dp_table and prefix_sum for the new values being computed.
6. Use nested loops for each potential highest value in the array and each potential cost. Calculate computation by considering the previous results:
   • Multiplying the count of configurations ending in highest_value with exactly curr_cost comparisons by highest_value.
   • Adding configurations possible with ending values less than highest_value and one less comparison count.
7. Update the current dp_table and keep the cumulative sums updated in prefix_sum.
8. After all iterations, the value of prefix_sum[max_val][cost] gives the total configurations for the specified conditions.

This model handles various scenarios effectively using the principles of dynamic programming and prefix sum arrays, considering the constraints given by the modulus operation for computational efficiency in a problem that has potential to operate on large numbers. The approach ensures that time complexity is manageable by careful recalculations only when necessary, made efficient by leveraging computed data from the previous iterations.