Tabulation Bottom Up DP in DSA Java - Time & Space Complexity

Time Complexity: Tabulation Bottom Up DP

O(n)

Understanding Time Complexity

We want to understand how the time needed grows when using tabulation in bottom-up dynamic programming.

Specifically, how the number of steps changes as the input size increases.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int fib(int n) {
  int[] dp = new int[n + 1];
  dp[0] = 0;
  dp[1] = 1;
  for (int i = 2; i <= n; i++) {
    dp[i] = dp[i - 1] + dp[i - 2];
  }
  return dp[n];
}
    

This code calculates the nth Fibonacci number using bottom-up tabulation.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop that fills the dp array from 2 to n.
• How many times: It runs (n - 1) times, which grows as n grows.

How Execution Grows With Input

Each step in the loop does a small fixed amount of work, so total work grows directly with n.

Input Size (n)	Approx. Operations
10	About 9 steps
100	About 99 steps
1000	About 999 steps

Pattern observation: The number of steps increases roughly in a straight line as n grows.

Final Time Complexity

Time Complexity: O(n)

This means the time needed grows directly in proportion to the input size.

Common Mistake

[X] Wrong: "Tabulation is always slower because it uses a loop over all n elements."

[OK] Correct: The loop does simple work each time, avoiding repeated calculations, so it is efficient and faster than naive recursion.

Interview Connect

Understanding this time complexity helps you explain why bottom-up DP is efficient and preferred in many problems.

Self-Check

"What if we changed the loop to run from 0 to n twice nested? How would the time complexity change?"

Practice

1. What is the main idea behind Tabulation Bottom Up DP in dynamic programming?

easy

A. Solve the problem by recursion without storing results

B. Start solving from the biggest problem and break it down

C. Use random guesses to find the solution

D. Build a table from the smallest subproblems up to the main problem

Solution

1. Step 1: Understand Tabulation Approach

   Tabulation means filling a table starting from the smallest subproblems to bigger ones.

2. Step 2: Compare with other methods

   Unlike recursion, tabulation avoids repeated work by storing results in a table.

3. Final Answer:

   Build a table from the smallest subproblems up to the main problem -> Option D

4. Quick Check:

   Tabulation = Bottom Up DP [OK]

Hint: Think bottom-up: solve small parts first, then bigger [OK]

Common Mistakes:

• Confusing tabulation with recursion
• Thinking tabulation starts from the biggest problem
• Ignoring the use of a table to store results

2. Which of the following is the correct way to initialize a DP array for bottom-up Fibonacci in Java?

easy

A. int[] dp = new int[n+1]; dp[0] = 0; dp[1] = 1;

B. int[] dp = new int[n]; dp[1] = 0; dp[2] = 1;

C. int[] dp = new int[n+1]; dp[0] = 1; dp[1] = 1;

D. int[] dp = new int[n]; dp[0] = 0; dp[1] = 1;

Solution

1. Step 1: Check array size for Fibonacci

   We need size n+1 to store from 0 to n indices.

2. Step 2: Set base cases correctly

   Fibonacci starts with dp[0]=0 and dp[1]=1.

3. Final Answer:

   int[] dp = new int[n+1]; dp[0] = 0; dp[1] = 1; -> Option A

4. Quick Check:

   Base cases dp[0]=0, dp[1]=1 [OK]

Hint: Array size n+1 and dp[0]=0, dp[1]=1 for Fibonacci [OK]

Common Mistakes:

• Using wrong array size causing IndexOutOfBounds
• Swapping base case values
• Starting indices from 1 instead of 0

3. What will be the output of the following Java code using bottom-up DP to find Fibonacci of 5?

int n = 5;
int[] dp = new int[n+1];
dp[0] = 0; dp[1] = 1;
for (int i = 2; i <= n; i++) {
    dp[i] = dp[i-1] + dp[i-2];
}
System.out.println(dp[n]);

medium

A. 13

B. 5

C. 3

D. 8

Solution

1. Step 1: Trace DP array filling

   dp[2]=1, dp[3]=2, dp[4]=3, dp[5]=5 by adding previous two values.

2. Step 2: Output dp[5]

   dp[5] is 5, so printed output is 5.

3. Final Answer:

   5 -> Option B

4. Quick Check:

   Fibonacci(5) = 5 [OK]

Hint: Remember Fibonacci sequence: 0,1,1,2,3,5... [OK]

Common Mistakes:

• Confusing index with Fibonacci value
• Off-by-one errors in loop
• Printing wrong dp index

4. Identify the error in this bottom-up DP code for Fibonacci:

int n = 5;
int[] dp = new int[n];
dp[0] = 0; dp[1] = 1;
for (int i = 2; i <= n; i++) {
    dp[i] = dp[i-1] + dp[i-2];
}
System.out.println(dp[n]);

medium

A. Array size is too small, causing IndexOutOfBoundsException

B. Base cases dp[0] and dp[1] are incorrect

C. Loop condition should be i < n instead of i <= n

D. No error, code runs fine

Solution

1. Step 1: Check array size

   Array dp has size n=5, valid indices 0 to 4 only.

2. Step 2: Loop accesses dp[5]

   Loop runs i=2 to 5 inclusive, dp[5] access causes error.

3. Final Answer:

   Array size is too small, causing IndexOutOfBoundsException -> Option A

4. Quick Check:

   Array size must be n+1 for dp[n] access [OK]

Hint: Array size must cover all indices used in loop [OK]

Common Mistakes:

• Using array size n instead of n+1
• Ignoring loop boundary causing runtime error
• Assuming code runs without error

5. You want to find the number of ways to climb n stairs if you can take 1 or 2 steps at a time using bottom-up DP. Which code snippet correctly implements this?

hard

A. int[] dp = new int[n+1]; dp[0]=0; dp[1]=1; for(int i=2;i<=n;i++){ dp[i]=dp[i-1]+dp[i-2]; } return dp[n];

B. int[] dp = new int[n]; dp[0]=1; dp[1]=2; for(int i=2;i

C. int[] dp = new int[n+1]; dp[0]=1; dp[1]=1; for(int i=2;i<=n;i++){ dp[i]=dp[i-1]+dp[i-2]; } return dp[n];

D. int[] dp = new int[n+1]; dp[0]=1; dp[1]=1; for(int i=1;i<=n;i++){ dp[i]=dp[i-1]+dp[i-2]; } return dp[n];

Solution

1. Step 1: Initialize base cases for climbing stairs

   dp[0]=1 means 1 way to stand still; dp[1]=1 means 1 way to climb 1 step.

2. Step 2: Fill dp using sum of previous two steps

   For i≥2, dp[i] = dp[i-1] + dp[i-2] counts ways taking 1 or 2 steps.

3. Step 3: Check loop and return

   Loop runs i=2 to n inclusive; return dp[n] is correct.

4. Final Answer:

   int[] dp = new int[n+1]; dp[0]=1; dp[1]=1; for(int i=2;i<=n;i++){ dp[i]=dp[i-1]+dp[i-2]; } return dp[n]; -> Option C

5. Quick Check:

   Climbing stairs DP uses dp[i]=dp[i-1]+dp[i-2] [OK]

Hint: Base dp[0]=1, dp[1]=1; sum previous two for dp[i] [OK]

Common Mistakes:

• Wrong base case values
• Multiplying instead of adding
• Incorrect loop start or end indices