Knight Dialer

class Solution {
public:
    int dialKnight(int steps) {
        if (steps == 1) {
            return 10;
        }
        
        int posA = 4, posB = 2, posC = 2, posD = 1;
        int MODULO = 1e9 + 7;
        
        for (int step = 0; step < steps - 1; step++) {
            int nextA = posA, nextB = posB, nextC = posC, nextD = posD;
            
            posA = ((2 * nextB) % MODULO + (2 * nextC) % MODULO) % MODULO;
            posB = nextA;
            posC = (nextA + (2 * nextD) % MODULO) % MODULO;
            posD = nextC;
        }
        
        int res = (posA + posB) % MODULO;
        res = (res + posC) % MODULO;
        return (res + posD) % MODULO;
    }
};

The "Knight Dialer" problem involves calculating the number of unique phone numbers that can be dialed using knight moves on a standard keypad. Your task is to compute the possibilities using a dynamic programming approach in C++.

• Start by initializing the phone pad positions posA, posB, posC, and posD representing the numbers of possible sequences ending at different groups of keys.

• Use a modulo constant MODULO for large numbers as 1e9 + 7 to prevent overflow.

• Use a loop to iterate through the required number of steps minus one, recalculating the position values based on the knight's move possibilities:

  • posA will update based on the current positions at posB and posC.
  • posB takes the value from the previous posA.
  • posC updates considering posD and posA.
  • posD takes the value from the previous posC.

• After iterating through the steps, combine the results from all ending positions (posA, posB, posC, posD) using modulo arithmetic to get the total number of sequences.

• Return the result which is the total count of all unique dialable numbers.

Apply these concepts in a method dialKnight that handles the computation for a given number of steps. The result is efficiently calculated using transitions derived from possible knight moves on the keypad.

class Solution {
    public int dialerKnight(int steps) {
        if (steps == 1) {
            return 10;
        }
        
        int X = 4;
        int Y = 2;
        int Z = 2;
        int W = 1;
        int MODULO = (int) 1e9 + 7;
        
        for (int j = 0; j < steps - 1; j++) {
            int newX = X;
            int newY = Y;
            int newZ = Z;
            int newW = W;
            
            X = ((2 * newY) % MODULO + (2 * newZ) % MODULO) % MODULO;
            Y = newX;
            Z = (newX + (2 * newW) % MODULO) % MODULO;
            W = newZ;
        }
        
        int result = (X + Y) % MODULO;
        result = (result + Z) % MODULO;
        return (result + W) % MODULO;
    }
}

The "Knight Dialer" algorithm calculates the number of different phone numbers that can potentially be dialed using a knight chess move pattern on a standard 3x4 number pad. Here's a concise breakdown of how the solution in Java works:

• Initialize variables for various counters (X, Y, Z, W) corresponding to the position of a knight after making certain moves.
• Use a loop controlled by given input steps. For each iteration, calculate new positions based on the knight's possible moves and update the counters.
• Use modulus operation with MODULO = 1e9 + 7 (common in algorithm challenges to prevent overflow and maintain numbers manageable).
• Update the counters (X, Y, Z, W) considering potential symmetries and overlapping moves on a typical phone keypad layout.
• At the end of all iterations, aggregate the counters modulo 1e9 + 7 to get the final number of unique numbers that can be dialed.

Ensure proper initialization before the loop begins to cover edge cases (here for steps == 1), resulting in an immediate return value of 10, representing one starting step from each dial pad number without any moves. The program iteratively handles transitions of knight moves and accumulates the result efficiently using simple arithmetic operations and modulus to prevent integer overflow.

class Dialer:
    def calculateMoves(self, steps: int) -> int:
        if steps == 1:
            return 10
        
        countA = 4
        countB = 2
        countC = 2
        countD = 1
        MODULO = 10 ** 9 + 7
        
        for i in range(steps - 1):
            countA, countB, countC, countD = (2 * (countB + countC)) % MODULO, countA, (countA + 2 * countD) % MODULO, countC
        
        return (countA + countB + countC + countD) % MODULO

The Knight Dialer problem involves calculating the number of distinct phone numbers of given length that can be dialed using a keypad, where the movement of the "knight" follows the rules similar to the knight in chess. Understanding your Python solution to this problem requires knowledge of dynamic programming and modular arithmetic.

• Initialize the function calculateMoves that takes steps—the length of the phone number—as an input parameter.
• Use base case handling such that if steps equals 1, the function returns 10. This indicates that from any number on the dial pad, you can dial any of the 10 digits.
• Set variables countA, countB, countC, and countD to represent different start positions on the keypad that have varying degrees of freedom to move. Their initial values correspond to their respective points on a typical dial pad accessible by the knight.
• Define MODULO as (10**9 + 7) to ensure all results are computed under this modulo, a common technique to avoid large number computations.
• Employ a for loop to iterate from the second step up to steps - 1 and recalculate counts countA, countB, countC, and countD based on prior values. This process accounts for how the knight moves from one key to another, adjusting for the number of unique numbers that can be produced.
• After completing the iterations, sum up all counts and take modulo 10**9 + 7 of the result to get the final count of the distinct phone numbers that can be dialed.

By the time the computation ends, calculateMoves returns the total number of unique dialable numbers for steps number of movements, taking modulo 10**9 + 7 with the current state of counts.

This solution efficiently uses dynamic programming to simplify a potentially complex computational problem by breaking it down into simpler subproblems with optimized calculations through each loop iteration.