
Problem Statement
In the problem, you are tasked with determining the number of distinct paths that a pointer can take to return to its starting position (index 0
) within a given number of moves, referred to as steps
. This pointer initiates at index 0
within an array whose length is defined by arrLen
. Throughout these steps, the pointer can either move one position to the right, one position to the left, or it can choose to remain stationary. However, it is crucial that the pointer never exits the boundaries of the array at any stage. To obtain the final count of such unique paths that allow the pointer to be back at index 0
after precisely steps
steps, the computation must be taken modulo 10^9 + 7
due to potential large size of the output.
Examples
Example 1
Input:
steps = 3, arrLen = 2
Output:
4
Explanation:
There are 4 differents ways to stay at index 0 after 3 steps. Right, Left, Stay Stay, Right, Left Right, Stay, Left Stay, Stay, Stay
Example 2
Input:
steps = 2, arrLen = 4
Output:
2
Explanation:
There are 2 differents ways to stay at index 0 after 2 steps Right, Left Stay, Stay
Example 3
Input:
steps = 4, arrLen = 2
Output:
8
Constraints
1 <= steps <= 500
1 <= arrLen <= 106
Approach and Intuition
In order to comprehend the approach for solving this problem, it's beneficial to dissect the problem based on provided examples and constraints.
Understanding the Dynamics through Examples:
Example 1:
steps = 3
,arrLen = 2
- The output is
4
; here, the limited length of the array and steps suggest that the pointer has a confined space to move: either stay at0
, move to1
and then come back, or use a combination of staying and moving. - This example explicitly emphasizes the basic movements: Right, Left, and Stay.
- The output is
Example 2:
steps = 2
,arrLen = 4
- Output is
2
; notably here, despite having a larger array length relative to the number of steps, the pointer can only move up to one position away and must return, or just stay at the initial position. LargerarrLen
more than steps by 2, doesn't provide more options due to the limitation by the number of steps.
- Output is
Example 3:
steps = 4
,arrLen = 2
- The output is
8
; with more steps available, combinations of staying and moving back-and-forth increase, explaining the higher number of distinct paths.
- The output is
Constraints Handling:
The maximum value for
steps
is500
, and that forarrLen
is1,000,000
. It's crucial to note that ifarrLen
is much greater thansteps
, only a part of the array equal tomin(arrLen, steps)
essentially affects the outcome since the pointer can't move beyond more positions than the number of steps available.Efficient calculation is paramount given the potential size of calculations. The result needs modulo
10^9 + 7
which hints at the implementation of dynamic programming to recursively determine the number of ways to stay at0
after a given number of moves, updating states based on previous computations.
Through these observations, one can imply that the solution entails a form of dynamic programming or memoization where each step's potential positions are calculated based on previous possibilities, while closely watching boundary conditions and efficiently managing large number computations using modulo operation.
Solutions
- C++
- Java
- Python
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
andmaxLen
.maxLen
is adjusted not to exceedtotalSteps
because it's impossible to move more steps away than the total steps available. - Two vectors,
curr
andlast
, are initialized for dynamic programming. They store the number of ways to be at each position from 0 tomaxLen - 1
after each step. - The vector
last
is initially set withlast[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 with1000000007
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 andmaxSteps
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, andprevious
, to hold values from the last computed step withprevious[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
from0
tomaxArrLen - 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 setprevious
tocurrent
for the next iteration.
- Reinitialize
- 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
andprevious
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.
- Sum up ways coming from the position itself and its immediate left and right neighbors, taking modulo
- Update the
previous
array with the values fromcurrent
for use in the next iteration.
- For each step, reset the
- 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.
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