Solution Approach
Solution Overview
There are 3 main approaches to solve this problem:
- Dynamic Programming (2D) - Complex approach
- Dynamic Programming (1D) - Complex approach
- Dynamic Programming with Further Optimization - Complex approach
Approach 1: Dynamic Programming (2D)
The key insight for solving this problem is to use dynamic programming to build up the solution from smaller subproblems.
Let's define dp[i][j] as the number of ways to be at position j after i steps.
For each step i and position j, we can consider three possible moves:
- Stay at position
j:dp[i-1][j]ways - Move left from position
j+1:dp[i-1][j+1]ways - Move right from position
j-1:dp[i-1][j-1]ways
Therefore, the recurrence relation is:
dp[i][j] = dp[i-1][j] + dp[i-1][j-1] + dp[i-1][j+1]
We need to be careful about the boundary conditions:
- If
j-1 < 0, thendp[i-1][j-1] = 0(can't move right from a negative position) - If
j+1 >= arrLen, thendp[i-1][j+1] = 0(can't move left from a position outside the array)
The base case is dp[0][0] = 1, as there is one way to be at position 0 after 0 steps.
The final answer is dp[steps][0], which represents the number of ways to be at position 0 after steps steps.
An important optimization is to observe that we only need to consider positions up to min(arrLen-1, steps), as we can't go further than steps positions away from the starting point. This is because each step can move us at most 1 position, so after steps steps, we can be at most steps positions away from the starting point.
Algorithm:
- Define maxPosition = min(arrLen - 1, steps), as we can't go further than steps positions away from the starting point.
- Initialize a 2D DP array dp of size (steps + 1) × (maxPosition + 1), with all elements set to 0.
- Set the base case: dp[0][0] = 1 (one way to be at position 0 after 0 steps).
- For each step i from 1 to steps:
- a. For each position j from 0 to maxPosition:
- i. Add dp[i-1][j] to dp[i][j] (stay at position j).
- ii. If j > 0, add dp[i-1][j-1] to dp[i][j] (move right from position j-1).
- iii. If j < maxPosition, add dp[i-1][j+1] to dp[i][j] (move left from position j+1).
- iv. Take modulo 10^9 + 7 to avoid integer overflow.
- Return dp[steps][0], which represents the number of ways to be at position 0 after steps steps.
Time Complexity:
We need to fill in the DP table of size (steps + 1) × (min(arrLen, steps) + 1), and each cell takes constant time to compute.
Space Complexity:
We use a 2D DP array of size (steps + 1) × (min(arrLen, steps) + 1) to store the number of ways for each step and position.
Approach 2: Dynamic Programming (1D)
We can optimize the space complexity of the dynamic programming approach by using a 1D array instead of a 2D array.
The key observation is that when we're calculating dp[i][j], we only need the values from the previous step dp[i-1][...]. So, we can use two 1D arrays, curr and prev, to store the current and previous steps' values, respectively.
The recurrence relation remains the same:
curr[j] = prev[j] + prev[j-1] + prev[j+1]
After each step, we swap curr and prev to prepare for the next step.
This approach reduces the space complexity from O(steps * min(arrLen, steps)) to O(min(arrLen, steps)), while maintaining the same time complexity.
Algorithm:
- Define maxPosition = min(arrLen - 1, steps), as we can't go further than steps positions away from the starting point.
- Initialize two 1D arrays prev and curr of size maxPosition + 1, with all elements set to 0.
- Set the base case: prev[0] = 1 (one way to be at position 0 after 0 steps).
- For each step i from 1 to steps:
- a. Reset curr to all 0s.
- b. For each position j from 0 to maxPosition:
- i. Add prev[j] to curr[j] (stay at position j).
- ii. If j > 0, add prev[j-1] to curr[j] (move right from position j-1).
- iii. If j < maxPosition, add prev[j+1] to curr[j] (move left from position j+1).
- iv. Take modulo 10^9 + 7 to avoid integer overflow.
- c. Swap prev and curr to prepare for the next step.
- Return prev[0], which represents the number of ways to be at position 0 after steps steps.
Time Complexity:
We still need to process each step and position, which takes O(steps * min(arrLen, steps)) time.
Space Complexity:
We only use two 1D arrays of size min(arrLen, steps) + 1 to store the number of ways for the current and previous steps.
Approach 3: Dynamic Programming with Further Optimization
We can further optimize the solution by observing that we don't need to consider all positions up to min(arrLen-1, steps). In fact, we only need to consider positions up to min(arrLen-1, steps/2).
This is because to return to position 0 after steps steps, we can go at most steps/2 positions away from the starting point. If we go further, we won't have enough steps to return to position 0.
For example, if steps = 10, we can go at most 5 positions away from the starting point (position 0), because we need 5 steps to go there and 5 steps to return.
This optimization further reduces the time and space complexity from O(steps * min(arrLen, steps)) to O(steps * min(arrLen, steps/2)).
Algorithm:
- Define maxPosition = min(arrLen - 1, steps / 2), as we can't go further than steps/2 positions away from the starting point if we want to return to position 0.
- Initialize two 1D arrays prev and curr of size maxPosition + 1, with all elements set to 0.
- Set the base case: prev[0] = 1 (one way to be at position 0 after 0 steps).
- For each step i from 1 to steps:
- a. Reset curr to all 0s.
- b. For each position j from 0 to maxPosition:
- i. Add prev[j] to curr[j] (stay at position j).
- ii. If j > 0, add prev[j-1] to curr[j] (move right from position j-1).
- iii. If j < maxPosition, add prev[j+1] to curr[j] (move left from position j+1).
- iv. Take modulo 10^9 + 7 to avoid integer overflow.
- c. Swap prev and curr to prepare for the next step.
- Return prev[0], which represents the number of ways to be at position 0 after steps steps.
Time Complexity:
We only need to process positions up to min(arrLen, steps/2), which reduces the time complexity.
Space Complexity:
We only use two 1D arrays of size min(arrLen, steps/2) + 1 to store the number of ways for the current and previous steps.
Pseudocode
12345678910111213141516171819202122232425function numWays(steps, arrLen): // Define the maximum position we need to consider maxPosition = min(arrLen - 1, steps) // Initialize DP array dp = new 2D array of size (steps + 1) × (maxPosition + 1), initialized with 0 // Base case dp[0][0] = 1 // Fill DP array for i from 1 to steps: for j from 0 to maxPosition: // Stay at position j dp[i][j] = dp[i-1][j] // Move right from position j-1 if j {'>'} 0: dp[i][j] = (dp[i][j] + dp[i-1][j-1]) % MOD // Move left from position j+1 if j {'<'} maxPosition: dp[i][j] = (dp[i][j] + dp[i-1][j+1]) % MOD return dp[steps][0]String Problems
Learning Objective
Understand different approaches to solve the number of ways to stay in the same place after some steps problem and analyze their efficiency.
Solution Approaches
Approach 1: Dynamic Programming (2D)
The key insight for solving this problem is to use dynamic programming to build up the solution from smaller subproblems.
Let's define dp[i][j] as the number of ways to be at position j after i steps.
For each step i and position j, we can consider three possible moves:
- Stay at position
j:dp[i-1][j]ways - Move left from position
j+1:dp[i-1][j+1]ways - Move right from position
j-1:dp[i-1][j-1]ways
Therefore, the recurrence relation is:
dp[i][j] = dp[i-1][j] + dp[i-1][j-1] + dp[i-1][j+1]
We need to be careful about the boundary conditions:
- If
j-1 < 0, thendp[i-1][j-1] = 0(can't move right from a negative position) - If
j+1 >= arrLen, thendp[i-1][j+1] = 0(can't move left from a position outside the array)
The base case is dp[0][0] = 1, as there is one way to be at position 0 after 0 steps.
The final answer is dp[steps][0], which represents the number of ways to be at position 0 after steps steps.
An important optimization is to observe that we only need to consider positions up to min(arrLen-1, steps), as we can't go further than steps positions away from the starting point. This is because each step can move us at most 1 position, so after steps steps, we can be at most steps positions away from the starting point.
Algorithm:
- Define maxPosition = min(arrLen - 1, steps), as we can't go further than steps positions away from the starting point.
- Initialize a 2D DP array dp of size (steps + 1) × (maxPosition + 1), with all elements set to 0.
- Set the base case: dp[0][0] = 1 (one way to be at position 0 after 0 steps).
- For each step i from 1 to steps:
- a. For each position j from 0 to maxPosition:
- i. Add dp[i-1][j] to dp[i][j] (stay at position j).
- ii. If j > 0, add dp[i-1][j-1] to dp[i][j] (move right from position j-1).
- iii. If j < maxPosition, add dp[i-1][j+1] to dp[i][j] (move left from position j+1).
- iv. Take modulo 10^9 + 7 to avoid integer overflow.
- Return dp[steps][0], which represents the number of ways to be at position 0 after steps steps.
Approach 2: Dynamic Programming (1D)
We can optimize the space complexity of the dynamic programming approach by using a 1D array instead of a 2D array.
The key observation is that when we're calculating dp[i][j], we only need the values from the previous step dp[i-1][...]. So, we can use two 1D arrays, curr and prev, to store the current and previous steps' values, respectively.
The recurrence relation remains the same:
curr[j] = prev[j] + prev[j-1] + prev[j+1]
After each step, we swap curr and prev to prepare for the next step.
This approach reduces the space complexity from O(steps * min(arrLen, steps)) to O(min(arrLen, steps)), while maintaining the same time complexity.
Algorithm:
- Define maxPosition = min(arrLen - 1, steps), as we can't go further than steps positions away from the starting point.
- Initialize two 1D arrays prev and curr of size maxPosition + 1, with all elements set to 0.
- Set the base case: prev[0] = 1 (one way to be at position 0 after 0 steps).
- For each step i from 1 to steps:
- a. Reset curr to all 0s.
- b. For each position j from 0 to maxPosition:
- i. Add prev[j] to curr[j] (stay at position j).
- ii. If j > 0, add prev[j-1] to curr[j] (move right from position j-1).
- iii. If j < maxPosition, add prev[j+1] to curr[j] (move left from position j+1).
- iv. Take modulo 10^9 + 7 to avoid integer overflow.
- c. Swap prev and curr to prepare for the next step.
- Return prev[0], which represents the number of ways to be at position 0 after steps steps.
Approach 3: Dynamic Programming with Further Optimization
We can further optimize the solution by observing that we don't need to consider all positions up to min(arrLen-1, steps). In fact, we only need to consider positions up to min(arrLen-1, steps/2).
This is because to return to position 0 after steps steps, we can go at most steps/2 positions away from the starting point. If we go further, we won't have enough steps to return to position 0.
For example, if steps = 10, we can go at most 5 positions away from the starting point (position 0), because we need 5 steps to go there and 5 steps to return.
This optimization further reduces the time and space complexity from O(steps * min(arrLen, steps)) to O(steps * min(arrLen, steps/2)).
Algorithm:
- Define maxPosition = min(arrLen - 1, steps / 2), as we can't go further than steps/2 positions away from the starting point if we want to return to position 0.
- Initialize two 1D arrays prev and curr of size maxPosition + 1, with all elements set to 0.
- Set the base case: prev[0] = 1 (one way to be at position 0 after 0 steps).
- For each step i from 1 to steps:
- a. Reset curr to all 0s.
- b. For each position j from 0 to maxPosition:
- i. Add prev[j] to curr[j] (stay at position j).
- ii. If j > 0, add prev[j-1] to curr[j] (move right from position j-1).
- iii. If j < maxPosition, add prev[j+1] to curr[j] (move left from position j+1).
- iv. Take modulo 10^9 + 7 to avoid integer overflow.
- c. Swap prev and curr to prepare for the next step.
- Return prev[0], which represents the number of ways to be at position 0 after steps steps.
Complexity Analysis
Dynamic Programming (2D) Approach
Time Complexity
We need to fill in the DP table of size (steps + 1) × (min(arrLen, steps) + 1), and each cell takes constant time to compute.
Space Complexity
We use a 2D DP array of size (steps + 1) × (min(arrLen, steps) + 1) to store the number of ways for each step and position.
Dynamic Programming (1D) Approach
Time Complexity
We still need to process each step and position, which takes O(steps * min(arrLen, steps)) time.
Space Complexity
We only use two 1D arrays of size min(arrLen, steps) + 1 to store the number of ways for the current and previous steps.
Dynamic Programming with Further Optimization Approach
Time Complexity
We only need to process positions up to min(arrLen, steps/2), which reduces the time complexity.
Space Complexity
We only use two 1D arrays of size min(arrLen, steps/2) + 1 to store the number of ways for the current and previous steps.
Pseudocode
Dynamic Programming (2D) Approach
12345678910111213141516171819202122232425function numWays(steps, arrLen): // Define the maximum position we need to consider maxPosition = min(arrLen - 1, steps) // Initialize DP array dp = new 2D array of size (steps + 1) × (maxPosition + 1), initialized with 0 // Base case dp[0][0] = 1 // Fill DP array for i from 1 to steps: for j from 0 to maxPosition: // Stay at position j dp[i][j] = dp[i-1][j] // Move right from position j-1 if j {'>'} 0: dp[i][j] = (dp[i][j] + dp[i-1][j-1]) % MOD // Move left from position j+1 if j {'<'} maxPosition: dp[i][j] = (dp[i][j] + dp[i-1][j+1]) % MOD return dp[steps][0]Dynamic Programming (1D) Approach
1234567891011121314151617181920212223242526272829303132function numWays(steps, arrLen): // Define the maximum position we need to consider maxPosition = min(arrLen - 1, steps) // Initialize arrays for previous and current steps prev = new array of size maxPosition + 1, initialized with 0 curr = new array of size maxPosition + 1, initialized with 0 // Base case prev[0] = 1 // Fill arrays for i from 1 to steps: // Reset current array curr = new array of size maxPosition + 1, initialized with 0 for j from 0 to maxPosition: // Stay at position j curr[j] = prev[j] // Move right from position j-1 if j {'>'} 0: curr[j] = (curr[j] + prev[j-1]) % MOD // Move left from position j+1 if j {'<'} maxPosition: curr[j] = (curr[j] + prev[j+1]) % MOD // Swap arrays for next step prev = curr return prev[0]Dynamic Programming with Further Optimization Approach
1234567891011121314151617181920212223242526272829303132function numWays(steps, arrLen): // Define the maximum position we need to consider maxPosition = min(arrLen - 1, steps / 2) // Initialize arrays for previous and current steps prev = new array of size maxPosition + 1, initialized with 0 curr = new array of size maxPosition + 1, initialized with 0 // Base case prev[0] = 1 // Fill arrays for i from 1 to steps: // Reset current array curr = new array of size maxPosition + 1, initialized with 0 for j from 0 to maxPosition: // Stay at position j curr[j] = prev[j] // Move right from position j-1 if j {'>'} 0: curr[j] = (curr[j] + prev[j-1]) % MOD // Move left from position j+1 if j {'<'} maxPosition: curr[j] = (curr[j] + prev[j+1]) % MOD // Swap arrays for next step prev = curr return prev[0]Dynamic Programming with Further Optimization Approach
1234567891011121314151617181920212223242526272829303132function numWays(steps, arrLen): // Define the maximum position we need to consider maxPosition = min(arrLen - 1, steps / 2) // Initialize arrays for previous and current steps prev = new array of size maxPosition + 1, initialized with 0 curr = new array of size maxPosition + 1, initialized with 0 // Base case prev[0] = 1 // Fill arrays for i from 1 to steps: // Reset current array curr = new array of size maxPosition + 1, initialized with 0 for j from 0 to maxPosition: // Stay at position j curr[j] = prev[j] // Move right from position j-1 if j {'>'} 0: curr[j] = (curr[j] + prev[j-1]) % MOD // Move left from position j+1 if j {'<'} maxPosition: curr[j] = (curr[j] + prev[j+1]) % MOD // Swap arrays for next step prev = curr return prev[0]