Number of Ways to Stay in the Same Place After Some Steps

class Solution {
public:
    int countWays(int totalSteps, int maxLen) {
        int modulo = 1000000007;
        maxLen = min(maxLen, totalSteps);
        vector<int> curr(maxLen, 0);
        vector<int> last(maxLen, 0);
        last[0] = 1;
            
        for (int step = 1; step <= totalSteps; step++) {
            curr = vector<int>(maxLen, 0);
                
            for (int pos = maxLen - 1; pos >= 0; pos--) {
                int sumWays = last[pos];
                    
                if (pos > 0) {
                    sumWays = (sumWays + last[pos - 1]) % modulo;
                }
                    
                if (pos < maxLen - 1) {
                    sumWays = (sumWays + last[pos + 1]) % modulo;
                }
                    
                curr[pos] = sumWays;
            }
                
            last = curr;
        }
            
        return curr[0];
    }
};

The provided C++ code offers a solution for determining the number of ways to remain at the starting position after performing a given number of steps (totalSteps), with a restriction on the maximum step length (maxLen). This problem resembles a dynamic programming challenge where the goal is to compute the ways staying at the initial position using the constraints provided. Here's a breakdown of how the solution operates:

• The function countWays receives two parameters: totalSteps and maxLen. maxLen is adjusted not to exceed totalSteps because it's impossible to move more steps away than the total steps available.
• Two vectors, curr and last, are initialized for dynamic programming. They store the number of ways to be at each position from 0 to maxLen - 1 after each step.
• The vector last is initially set with last[0] = 1, indicating one way to stay at the starting position before any moves.
• A nested loop iterates through each step and each position recalculating the ways to stay at each position based on the prior step. This recalculation considers:
  • Staying at the same position.
  • Moving to the position from the step immediately before it.
  • Returning from the position immediately after it, if any.
• Updates are done to the curr vector using modulo operation with 1000000007 to manage large number outputs and prevent overflow.
• After iterating through all steps, the solution returns the number of ways to be at the initial position (curr[0]), which is the result.

Use this structured approach to solve similar dynamic programming problems efficiently by focusing on how states transition between steps while considering constraints like maximum lengths or limits.

class Solution {
    public int countUniquePaths(int maxSteps, int maxArrLen) {
        int MODULO = (int) 1e9 + 7;
        maxArrLen = Math.min(maxArrLen, maxSteps);
        int[] current = new int[maxArrLen];
        int[] previous = new int[maxArrLen];
        previous[0] = 1;
            
        for (int step = 1; step <= maxSteps; step++) {
            current = new int[maxArrLen];
                
            for (int pos = maxArrLen - 1; pos >= 0; pos--) {
                int result = previous[pos];
                if (pos > 0) {
                    result = (result + previous[pos - 1]) % MODULO;
                }
                    
                if (pos < maxArrLen - 1) {
                    result = (result + previous[pos + 1]) % MODULO;
                }
                    
                current[pos] = result;
            }
                
            previous = current;
        }
            
        return current[0];
    }
}

This Java program tackles the problem of computing the number of distinct ways to stay in the same position after a given number of steps, where movement can be either to the left or right within a bounded array. The method countUniquePaths accomplishes this by employing dynamic programming.

• Define "MODULO" with a large prime to handle large numbers and avoid overflow.
• Adjust maxArrLen to be the minimum of itself and maxSteps for efficiency, since moving beyond half the number of steps outward from the center isn't possible.
• Initialize a dynamic programming array, current, which updates its values for every step, and previous, to hold values from the last computed step with previous[0] initialized to 1, signifying there's initially one way to be at the start.
• Iterate through each step up to maxSteps. For each step:
  • Reinitialize current to make sure previous step values don't interfere.
  • Calculate the number of ways to be at each position pos from 0 to maxArrLen - 1.
  • For each position, sum ways from the same position and its adjacent positions from the previous step, taking care of boundary conditions.
  • Update the current array with these results, then set previous to current for the next iteration.
• Finally, return the value at the starting position (current[0]), which indicates the number of ways one can end up at the original position after all steps.

The approach ensures efficient computation by optimizing space with only two arrays and minimizing redundant calculations. The solution also maintains performance with modular arithmetic to handle potentially large numbers, ensuring the solution's suitability for high constraints on step and array size.

class Solution:
    def calculateWays(self, totalSteps: int, length: int) -> int:
        MODULO = 10 ** 9 + 7
        length = min(length, totalSteps)
        current = [0] * (length)
        previous = [0] * (length)
        previous[0] = 1
            
        for step in range(1, totalSteps + 1):
            current = [0] * (length)
                
            for index in range(length - 1, -1, -1):
                result = previous[index]
                    
                if index > 0:
                    result = (result + previous[index - 1]) % MODULO
                    
                if index < length - 1:
                    result = (result + previous[index + 1]) % MODULO
                    
                current[index] = result
                    
            previous = current
            
        return current[0]

This Python program calculates the number of ways a person can stay in the same position after taking a specified number of steps, considering that after each step, the person can either move left, right, or stay in place. This problem is solved using dynamic programming, where the possible positions and the total number of steps constrain the computation.

• Initialize two arrays, current and previous to track the number of ways to end at each position for the current step and the previous step respectively.
• Set all positions in these arrays to zero except for the starting point, which is set to 1 in the previous array.
• Iterate over each step from 1 to the total number of steps.
  • For each step, reset the current array.
  • For each index position in the array (from the last index down to the first), compute the number of ways to reach that position:
    • Sum up ways coming from the position itself and its immediate left and right neighbors, taking modulo 1,000,000,007 to handle large numbers.
  • Update the previous array with the values from current for use in the next iteration.
• The answer, the number of ways to remain at the starting position after all steps, will be found in current[0].

This method efficiently computes the number of ways using O(totalSteps * length) time and space complexity, where length is limited to the minimum of the area size and totalSteps to optimize performance.