Perfect — let’s scale it up so it really “clicks”.

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📦 Bigger Example

s = "abcdefghijk"   // length n = 11
lengths = [10, 1, 4]

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🧠 Meaning of operations

Each L means:

roll indices [0 .. L-1] by +1

So:

• 10 → affects indices 0..9
• 1 → affects index 0
• 4 → affects indices 0..3

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Step 1 — Initialize diff

diff = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]   // size n+1 = 12

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Step 2 — Apply updates

🔹 L = 10

diff[0] += 1
diff[10] -= 1

diff = [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0]

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🔹 L = 1

diff[0] += 1
diff[1] -= 1

diff = [2, -1, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0]

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🔹 L = 4

diff[0] += 1
diff[4] -= 1

diff = [3, -1, 0, 0, -1, 0, 0, 0, 0, 0, -1, 0]

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⚡ Step 3 — Prefix sum (critical step)

We now compute how many times each index is rolled.

running = 0

i	diff[i]	running (roll count)
0	+3	3
1	-1	2
2	0	2
3	0	2
4	-1	1
5	0	1
6	0	1
7	0	1
8	0	1
9	0	1
10	-1	0

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✅ Final roll counts

[3, 2, 2, 2, 1, 1, 1, 1, 1, 1, 0]

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🔤 Step 4 — Apply to string

index:   0 1 2 3 4 5 6 7 8 9 10
char:    a b c d e f g h i j  k
rolls:   3 2 2 2 1 1 1 1 1 1  0

Apply shifts:

• a +3 → d
• b +2 → d
• c +2 → e
• d +2 → f
• e +1 → f
• f +1 → g
• g +1 → h
• h +1 → i
• i +1 → j
• j +1 → k
• k +0 → k

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🎯 Final result

"ddeffghijkk"

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🧠 What to remember (important)

👉 Each L adds:

+1 to [0 .. L-1]

👉 diff encodes:

start adding at 0
stop adding at L

👉 prefix sum converts that into:

actual roll count per index

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⚡ Intuition shortcut

Instead of thinking:

"for each index, check all lengths"

Think:

"each length paints a prefix — diff array stacks all paints efficiently"

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If you want, next step I can show:

👉 how to solve this without diff using sorting + binary search (O(n log n)) — sometimes interviewers push that variant.