Tabulation (bottom-up DP) — iterative — Hinglish notes

3.7.8 · Coding › Algorithm Paradigms

WHAT hai tabulation?

YEH KAAM KYU KARTA HAI? Ek DP problem mein optimal substructure hoti hai (answer sub-answers se banta hai) aur overlapping subproblems hote hain (same sub-answers baar baar chahiye hote hain). Agar hum har subproblem ko ek aise order mein ek baar compute karein jahan uski dependencies pehle se done hon, toh hum kabhi recompute nahi karte aur kabhi recurse nahi karte.

HOW banayein koi bhi tabulation (the recipe)

Canonical example derive karna: Fibonacci

Hame $F (n)$ chahiye jahan $F (0) = 0, F (1) = 1, F (n) = F (n - 1) + F (n - 2)$ .

Step 1 — State. dp[i] = $i$ -va Fibonacci number. Step 2 — Recurrence. Definition se $d p [i] = d p [i - 1] + d p [i - 2]$ . Yeh step kyu? Har Fibonacci number literally pehle wale do ka sum hota hai — yahi optimal substructure IS. Step 3 — Base cases. dp[0]=0, dp[1]=1. Step 4 — Order. Increasing $i$ : jab hum dp[i] compute karte hain, dp[i-1] aur dp[i-2] dono pehle se exist karte hain. ✅ Step 5 — Answer. dp[n].

def fib(n):
    if n < 2: return n
    dp = [0]*(n+1)
    dp[1] = 1
    for i in range(2, n+1):
        dp[i] = dp[i-1] + dp[i-2]   # read smaller, already filled
    return dp[n]

Figure — Tabulation (bottom-up DP) — iterative

Worked example 2 — Climbing Stairs (paths count karna)

Tum $n$ stairs chadh rahe ho, ek baar mein 1 ya 2 steps lete ho. Kitne distinct ways hain?

State. dp[i] = step $i$ tak pahunchne ke ways ki sankhya. Recurrence. Step $i$ pe land karne ke liye, aakhri move ya toh 1-step tha ( $i - 1$ se) ya 2-step ( $i - 2$ se): $d p [i] = d p [i - 1] + d p [i - 2] .$ Yeh step kyu? Ways ka set last action ke basis pe clearly split hota hai; dono cases disjoint hain, isliye hum unhe add karte hain. Base. dp[0]=1 (ek way: khadhe raho), dp[1]=1. Answer. dp[n].

def climb(n):
    dp = [0]*(n+1)
    dp[0] = dp[1] = 1
    for i in range(2, n+1):
        dp[i] = dp[i-1] + dp[i-2]
    return dp[n]

Fibonacci jaisi hi shape hai — bahut saare DP problems ek hi skeleton share karte hain.

Worked example 3 — 0/1 Knapsack (2D table)

Items hain weights $w_{i}$ , values $v_{i}$ ke saath, capacity $W$ hai. Value maximize karo, har item zyada se zyada ek baar use ho.

State. dp[i][c] = pehle $i$ items aur capacity $c$ ke saath best value. Recurrence. Item $i$ ke liye (1-indexed, weight $w_{i}$ , value $v_{i}$ ): $d p [i] [c] = ⎩ ⎨ ⎧ d p [i - 1] [c] max (skip d p [i - 1] [c], take v_{i} + d p [i - 1] [c - w_{i}]) if w_{i} > c (can’t take it) otherwise$ Yeh step kyu? Har item ya toh andar hai ya bahar — yahi binary choice substructure hai. Agar hum lete hain, capacity $w_{i}$ se ghatti hai aur value $v_{i}$ se badhti hai, ek chhota subproblem dp[i-1][c-w_i] bachta hai. Base. dp[0][c]=0 (koi items nahi → koi value nahi). Order. Increasing $i$ (rows), $c$ pe koi bhi order, kyunki har read row $i - 1$ use karta hai jo poori tarah done hai.

def knapsack(w, v, W):
    n = len(w)
    dp = [[0]*(W+1) for _ in range(n+1)]
    for i in range(1, n+1):
        for c in range(W+1):
            dp[i][c] = dp[i-1][c]                     # skip item i
            if w[i-1] <= c:
                dp[i][c] = max(dp[i][c],
                               v[i-1] + dp[i-1][c-w[i-1]])  # take it
    return dp[n][W]

Common mistakes (steel-manned)

Recall Feynman: ek 12-saal ke bacche ko explain karo

Socho LEGO tower banana hai. 4tha brick lage bina 5va nahi rakh sakte. Toh neeche se shuru karo: brick 1 rakhna, phir 2 brick 1 use karke, phir 3 brick 2 use karke… Har brick yaad rakhta hai neeche kya hai, isliye tum kabhi kuch dobara nahi banate. Tabulation exactly yahi hai: chhote answers boxes mein likhte jao, phir unke upar bade answers stack karte jao — ek loop, koi magic nahi, koi peeche jaana nahi.

Active recall

Tabulation DP ki kaun si direction hai?

Bottom-up — sabse chhote subproblems se shuru karo aur upar ki taraf iterate karo, results ek table mein store karte hue.

Tabulation recursion overhead kyu avoid karta hai?

Yeh har subproblem ko dependency order mein loops use karke ek baar compute karta hai; kuch recompute nahi hota aur koi call stack nahi hoti.

DP/tabulation ke liye ek problem mein kaunsi do properties honi chahiye?

Optimal substructure aur overlapping subproblems.

Table kis order mein fill karna chahiye?

Aise order mein jahan har entry ki dependencies pehle se computed hon (usually chhota index → bada index).

Fibonacci tabulation recurrence aur base cases?

dp[i]=dp[i-1]+dp[i-2]; dp[0]=0, dp[1]=1.

Climbing stairs recurrence — yeh max ki jagah sum kyu hai?

dp[i]=dp[i-1]+dp[i-2]; hum paths ke disjoint sets count kar rahe hain (aakhri step 1 ya 2 tha), isliye add karte hain.

0/1 knapsack take-vs-skip recurrence?

dp[i][c]=max(dp[i-1][c], v_i+dp[i-1][c-w_i]) jab w_i≤c, warna dp[i-1][c].

Space-optimized 1D knapsack mein capacity high→low kyu loop karte hain?

Taki dp[c-w] abhi bhi previous row (item reuse nahi hua) ko reference kare, 0/1 semantics preserve ho.

Memoization aur tabulation mein kya farq hai?

Memoization = top-down recursion + cache; tabulation = bottom-up iterative table filling. Same complexity, tabulation mein koi recursion cost nahi.

Tabulation table mein final answer usually kahan hota hai?

Sabse bade/last state cell mein, jaise dp[n] ya dp[n][W].

Connections

Memoization (top-down DP) — is note ka recursive dual
Dynamic Programming — parent paradigm
Optimal Substructure aur Overlapping Subproblems — prerequisites
0-1 Knapsack, Longest Common Subsequence, Coin Change — classic tabulation problems
Space Optimization in DP — 2D tables se rolling arrays
Recurrence Relations — recurrence step ke neeche ka math