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Understanding Unmet Needs

The Real Question​

Scenario:

  • freq[x] = 0 (no x available)
  • need[x] > 0 (some chain is waiting for x)

We have the guard:

if (freq[x] == 0) continue;

So we SKIP processing x.

BUT... what about the chain that's waiting for x?

Doesn't it need x to continue? Don't we FAIL in this case?

In other words: "If we skip x because freq[x] == 0, but need[x] > 0, isn't that a problem?"

The Answer: It's Actually NOT a Problem!​

This is subtle but beautiful. Unmet needs are perfectly fine!

When Does This Scenario Happen?​

For this scenario to occur, we need:

  • freq[x] = 0 (x was all used up OR never existed)
  • need[x] > 0 (a chain ending at x-1 wants x)

Path 1: x Was in the Array But Got Consumed​

Input: [1, 2, 3, 3, 4, 5]
              ↑   ↑
           first second

Process first 3:
  - Used in some chain
  - freq[3] = 2 → 1

Process second 3:
  - Used in some chain
  - freq[3] = 1 → 0

Later, if need[3] > 0:
  - freq[3] = 0
  - Guard skips it
  - need[3] remains unsatisfied

Path 2: x Never Existed in the Array​

Input: [1, 2, 4, 5, 6]  (no 3!)

Process 1, 2:
  - Maybe started a chain [1, 2, ...]
  - This creates need[3] = 1

Process 4, 5, 6...
  - 3 never appears in the loop!
  - need[3] = 1 remains unmet

The Critical Insight: When We Fail vs When We Don't​

We FAIL When: A Number Can't Be Used​

The algorithm fails when we encounter a number that can't go anywhere:

for (int x : nums) {
    if (freq[x] == 0) continue;

    if (need[x] > 0) {
        // EXTEND
    }
    else if (freq[x+1] > 0 && freq[x+2] > 0) {
        // START
    }
    else {
        return "fail";  // ← Number is stuck! Can't extend or start
    }
}

Failure happens at:

"I have x, but I can't extend with it, and I can't start with it"

NOT at:

"Some chain needs x, but x isn't available"

Example: Missing Number Causes Immediate Failure​

Input: [1, 2, 4, 5, 6]  (missing 3)

Process x=1:
  need[1] = 0 (no demand)
  Can START? Need [1, 2, 3]

  Check:
  - freq[2] = 1 ✓
  - freq[3] = 0 ✗  ← Missing!

  Can't extend (no demand)
  Can't start (missing 3)

  → FAIL immediately!

We catch the problem BEFORE creating need[3]!

The algorithm never allows us to create a chain that can't be completed.

Example: Running Out of Numbers​

Input: [1, 2, 3, 4, 4, 5]

Process 1: START [1, 2, 3]
  freq = {1:0, 2:0, 3:0, 4:2, 5:1}
  need = {4:1}

Process 2: skip (freq[2] = 0)
Process 3: skip (freq[3] = 0)

Process first 4:
  freq[4] = 2 ✓
  need[4] = 1 ✓

  EXTEND [1,2,3] → [1,2,3,4]
  freq = {4:1, 5:1}
  need = {5:1}

Process second 4:
  freq[4] = 1 ✓
  need[4] = 0 (no demand for 4)

  Can START? Need [4, 5, 6]
  - freq[5] = 1 ✓
  - freq[6] = 0 ✗

  Can't extend (no demand for 4)
  Can't start (missing 6)

  → FAIL! (Number 4 is stuck)

We fail at the point where a number can't be used!

Why Unmet Needs Are OK​

The Key Understanding​

When need[x] > 0, it means:

"There's a chain [..., x-2, x-1] that COULD continue with x if x is available"

NOT:

"There's a chain that MUST have x or else it's invalid"

The chain is already valid (≥ 3 length) when it creates the need!

Example: Valid Case with Unmet Needs​

Input: [1, 2, 3, 3, 4, 5]

Process 1: START [1, 2, 3]
  need = {4:1}

Process second 3: START [3, 4, 5]
  need = {4:1, 6:1}

Process 4:
  freq[4] = 1
  need[4] = 1

  EXTEND (satisfies [1,2,3]'s need)
  freq[4] = 0
  need = {5:1, 6:1}

  Note: [3,4,5] also needed 4, but we only had one!

Process 5:
  freq[5] = 1
  need[5] = 1

  EXTEND (satisfies [1,2,3,4]'s need)
  freq[5] = 0
  need = {6:2}

  Note: [3,4,5] also needed 5, but we only had one!

End of loop:
  need = {6:2}  ← Two unmet needs!

Are these chains valid?

Chain 1: [1, 2, 3, 4, 5]  length = 5 ✓
Chain 2: [3, 4, 5]        length = 3 ✓

Both are ≥ 3, so both are VALID!

The unmet need[6] = 2 means:

  • Both chains could continue if 6 existed
  • But they don't NEED to continue
  • They're already valid!

The Complete Picture​

When freq[x] = 0 and need[x] > 0:​

Case 1: x is in nums[] (appears in loop)​

if (freq[x] == 0) continue;
  • We skip this iteration
  • need[x] remains unsatisfied
  • This is OK! The chain waiting for x is already ≥ 3 length

Case 2: x is NOT in nums[] (never appears in loop)​

  • We never process x at all
  • need[x] remains unsatisfied
  • This is OK! The chain waiting for x is already ≥ 3 length

Case 3: We need x to START a chain​

  • We would have failed earlier!
  • At the number that tried to start
  • This prevents the problem from occurring

Visual: The Two Types of Unsatisfied Needs​

Type 1: Harmless (chain already valid)​

[1, 2, 3, ...] needs 4
      ↑
   Already length 3 ✓

If 4 never comes: Still valid!

Type 2: Harmful (chain too short)​

[1, 2, ...] needs 3
   ↑
  Only length 2 ✗

If 3 never comes: INVALID!

But Type 2 can't happen!

Why? Because we only create need[3] when we START a chain, and START requires reserving x, x+1, x+2:

else if (freq[x+1] > 0 && freq[x+2] > 0) {
    // Reserve x, x+1, x+2 (length 3 minimum guaranteed!)
    freq.put(x, freq.get(x) - 1);
    freq.put(x + 1, freq.get(x + 1) - 1);
    freq.put(x + 2, freq.get(x + 2) - 1);
    need.put(x + 3, ...);  // Need for x+3, not x+2!
}

We never create a need for a number that would make a chain too short!

The Only Failure Point​

The algorithm has exactly ONE place where it can fail:

else {
    return "fail";  // ← Only failure point
}

This happens when:

  1. We have a number x available (freq[x] > 0)
  2. But we can't use it anywhere:
    • Can't extend (no demand: need[x] == 0)
    • Can't start (missing x+1 or x+2)

This does NOT happen when:

  • We need x but don't have it
  • Some chain is waiting for x that never comes

Why This Design Is Correct​

Chains Are Created Valid​

  • START requires 3 numbers minimum: [x, x+1, x+2]
  • EXTEND adds to chains that are already valid
  • Impossible to have invalid chains

Needs Are Created For Extensions Only​

  • need[x+3] is created after forming [x, x+1, x+2]
  • The chain is already valid before creating the need
  • If x+3 never comes, the chain stays valid at length 3

Failures Happen At Source​

  • If we can't form a valid subsequence, we fail immediately
  • We don't fail later when needs go unmet
  • The "stuck number" is the real problem

Summary​

Your question: "What if freq[x] = 0 but need[x] > 0? We skip x, but doesn't that leave a chain hanging?"

Answer:

  1. The chain is already valid!

    • When we create need[x], the chain is already ≥ 3 length
    • It doesn't NEED x to be valid
    • It just COULD continue with x

  2. We only fail when a number is stuck

    • If we have x but can't use it anywhere
    • NOT when we need x but don't have it

  3. The algorithm prevents short chains

    • START requires x, x+1, x+2 (minimum length 3)
    • EXTEND adds to chains already ≥ 3
    • Impossible to have invalid chains

The beauty: Unmet needs are not failures! They just mean "chains that could have been longer but are already valid."