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$ curl repovive.com/roadmaps/greedy-algorithms/array-greedy-problems/partition-labels-implementation

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#####   ######  #####    ###    #   #  ###  #   #  ######
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$ curl repovive.com/roadmaps/greedy-algorithms/array-greedy-problems/partition-labels-implementation

Repovive

Partition Labels - Implementation - Array Greedy Problems | Greedy Algorithms | Repovive

Repovive.

Greedy Algorithms8 sections · 317 units14h total

Open in Course

Partition Labels - Implementation

The code

Here is the solution:

function partitionLabels(s)
 lastIndex := map from char to int
 for i from 0 to length - 1
 lastIndex[s[i]] := i
 result := []
 start := 0
 end := 0
 for i from 0 to length - 1
 end := max(end, lastIndex[s[i]])
 if i = end then
 result.add(end - start + 1)
 start := i + 1
 return result

Time: $O(n)$ . Space: $O(1)$ since alphabet is constant size.

When $i$ reaches $end$ , all characters in $[start, end]$ have their last occurrences within this range. Safe to cut here.