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DSA Typescriptprogramming~10 mins

Coin Change Total Number of Ways in DSA Typescript - Execution Trace

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Concept Flow - Coin Change Total Number of Ways
Start with empty ways array
Initialize ways[0
For each coin
For each amount from coin to target
Update ways[amount
Repeat for all coins
Result: ways[target
We build an array where each index shows ways to make that amount. For each coin, we add ways to make amounts including that coin.
Execution Sample
DSA Typescript
const coins = [1, 2, 3];
const amount = 4;
const ways = Array(amount + 1).fill(0);
ways[0] = 1;
for (const coin of coins) {
  for (let i = coin; i <= amount; i++) {
    ways[i] += ways[i - coin];
  }
}
Calculate total ways to make amount 4 using coins 1, 2, and 3.
Execution Table
StepOperationAmount Index Updatedways Array StateExplanation
0Initialize ways0[1, 0, 0, 0, 0]ways[0] = 1 means 1 way to make amount 0 (use no coins)
1Use coin 11[1, 1, 0, 0, 0]ways[1] += ways[0] → 0 + 1 = 1
2Use coin 12[1, 1, 1, 0, 0]ways[2] += ways[1] → 0 + 1 = 1
3Use coin 13[1, 1, 1, 1, 0]ways[3] += ways[2] → 0 + 1 = 1
4Use coin 14[1, 1, 1, 1, 1]ways[4] += ways[3] → 0 + 1 = 1
5Use coin 22[1, 1, 2, 1, 1]ways[2] += ways[0] → 1 + 1 = 2
6Use coin 23[1, 1, 2, 2, 1]ways[3] += ways[1] → 1 + 1 = 2
7Use coin 24[1, 1, 2, 2, 3]ways[4] += ways[2] → 1 + 2 = 3
8Use coin 33[1, 1, 2, 3, 3]ways[3] += ways[0] → 2 + 1 = 3
9Use coin 34[1, 1, 2, 3, 4]ways[4] += ways[1] → 3 + 1 = 4
10End-[1, 1, 2, 3, 4]All coins processed, ways[4] = 4 total ways
💡 All coins processed, final ways array shows total ways to make each amount
Variable Tracker
VariableStartAfter Step 1After Step 5After Step 8Final
ways[1,0,0,0,0][1,1,0,0,0][1,1,2,1,1][1,1,2,3,3][1,1,2,3,4]
coinN/A1233
i (amount index)N/A1234
Key Moments - 3 Insights
Why do we set ways[0] = 1 at the start?
ways[0] = 1 means there is exactly one way to make amount 0: by choosing no coins. This is the base case that allows building up solutions for larger amounts (see execution_table step 0).
Why do we add ways[i - coin] to ways[i] inside the loop?
Adding ways[i - coin] means we count all ways to make the smaller amount (i - coin) and then add the current coin to reach amount i. This accumulates all combinations including the current coin (see execution_table steps 1-9).
Why do we loop coins first, then amounts, not the other way?
Looping coins first ensures combinations are counted without duplicates. Each coin adds ways to amounts in order, so we don't count the same combination multiple times in different orders (see concept_flow).
Visual Quiz - 3 Questions
Test your understanding
Look at the execution_table at step 7, what is the value of ways[4]?
A1
B2
C3
D4
💡 Hint
Check the 'ways Array State' column at step 7 in execution_table
At which step does ways[3] first become 3?
AStep 8
BStep 6
CStep 3
DStep 9
💡 Hint
Look at the 'ways Array State' column for ways[3] in execution_table rows
If we remove coin 1 from the coins array, how would ways[4] change after processing all coins?
AIt would become 3
BIt would become 1
CIt would stay 4
DIt would become 0
💡 Hint
Without coin 1, fewer combinations exist; check how coin 1 updates ways in execution_table steps 1-4
Concept Snapshot
Coin Change Total Number of Ways:
- Use a ways array of size amount+1
- Initialize ways[0] = 1 (base case)
- For each coin, update ways[i] += ways[i - coin] for i >= coin
- ways[amount] gives total combinations
- Order of loops avoids duplicates
Full Transcript
This visualization shows how to find the total number of ways to make a target amount using given coin denominations. We start with an array 'ways' where ways[0] = 1, meaning one way to make zero amount. For each coin, we update the ways array by adding ways to make smaller amounts. The execution table traces each update step by step, showing how the array changes. Key moments clarify why ways[0] is set to 1, why we add ways[i - coin], and why the coin loop comes first. The quiz tests understanding of array states at specific steps and effects of removing coins. The snapshot summarizes the approach in simple steps.