Summary Ranges

▶

Steps

setup

Initialize pointers and ranges list

Initialize an empty list 'ranges' to store the summary ranges. Set 'start' pointer to 0 to begin scanning from the first element.

💡 Starting with pointers and an empty list sets the stage for tracking consecutive sequences.

Line:

ranges = []
n = len(nums)
start = 0

💡 We are ready to scan the array from the beginning to find consecutive ranges.

setup

Set 'end' pointer to 'start'

Set 'end' pointer equal to 'start' to begin checking for consecutive numbers starting at the current position.

💡 The 'end' pointer will move forward to find the end of the current consecutive range.

Line:

end = start

💡 'end' marks the current end of the range being explored.

compare

Check if next number is consecutive

Check if the number at 'end + 1' is exactly one greater than the number at 'end'. If yes, move 'end' forward to extend the range.

💡 This comparison detects if the current sequence continues consecutively.

Line:

while end + 1 < n and nums[end + 1] == nums[end] + 1:
    end += 1

💡 The sequence 0 -> 1 is consecutive, so we extend the range.

compare

Continue extending range for consecutive numbers

Check again if the next number after 'end' is consecutive. Since nums[2] == nums[1] + 1 (2 == 1 + 1), move 'end' forward to 2.

💡 Repeatedly checking consecutive condition extends the range to include all consecutive numbers.

Line:

while end + 1 < n and nums[end + 1] == nums[end] + 1:
    end += 1

💡 The sequence 0 -> 1 -> 2 is consecutive, so the range grows.

compare

Next number breaks consecutive sequence

Check if nums[3] == nums[2] + 1 (4 == 2 + 1). This is false, so the consecutive sequence ends at 'end' = 2.

💡 Detecting the break in sequence triggers recording the current range.

Line:

while end + 1 < n and nums[end + 1] == nums[end] + 1:
    end += 1

💡 The consecutive range from 0 to 2 is complete and ready to be recorded.

insert

Record the first range '0->2'

Since 'start' != 'end', append the range '0->2' to the 'ranges' list representing the consecutive numbers from index 0 to 2.

💡 Recording the range captures the consecutive sequence found so far.

Line:

if start == end:
    ranges.append(str(nums[start]))
else:
    ranges.append(f"{nums[start]}->{nums[end]}")

💡 Ranges are stored as strings representing the start and end of consecutive sequences.

shrink

Move 'start' pointer to next position after 'end'

Set 'start' to 'end + 1' (3) to begin searching for the next consecutive range starting at index 3.

💡 Advancing 'start' moves the scanning window forward to find new ranges.

Line:

start = end + 1

💡 'start' now points to the next unprocessed number.

setup

Set 'end' pointer to 'start' for next range

Set 'end' equal to 'start' (3) to begin checking for consecutive numbers starting at index 3.

💡 Resetting 'end' prepares to find the next consecutive range.

Line:

end = start

💡 'end' pointer resets to track the new range.

compare

Check if next number is consecutive for second range

Check if nums[4] == nums[3] + 1 (5 == 4 + 1). True, so move 'end' forward to 4 to extend the range.

💡 Detecting consecutive numbers extends the current range.

Line:

while end + 1 < n and nums[end + 1] == nums[end] + 1:
    end += 1

💡 The sequence 4 -> 5 is consecutive, so the range grows.

compare

Next number breaks consecutive sequence again

Check if nums[5] == nums[4] + 1 (7 == 5 + 1). False, so the consecutive sequence ends at 'end' = 4.

💡 Detecting the break triggers recording the current range.

Line:

while end + 1 < n and nums[end + 1] == nums[end] + 1:
    end += 1

💡 The consecutive range from 4 to 5 is complete and ready to be recorded.

insert

Record the second range '4->5'

Since 'start' != 'end', append the range '4->5' to the 'ranges' list representing the consecutive numbers from index 3 to 4.

💡 Recording the range captures the consecutive sequence found in the second segment.

Line:

if start == end:
    ranges.append(str(nums[start]))
else:
    ranges.append(f"{nums[start]}->{nums[end]}")

💡 Ranges are stored as strings representing the start and end of consecutive sequences.

shrink

Move 'start' pointer to next position after 'end'

Set 'start' to 'end + 1' (5) to begin searching for the next consecutive range starting at index 5.

💡 Advancing 'start' moves the scanning window forward to find new ranges.

Line:

start = end + 1

💡 'start' now points to the next unprocessed number.

setup

Set 'end' pointer to 'start' for last range

Set 'end' equal to 'start' (5) to begin checking for consecutive numbers starting at index 5.

💡 Resetting 'end' prepares to find the last consecutive range or single number.

Line:

end = start

💡 'end' pointer resets to track the new range.

compare

Check if next number is consecutive for last element

Check if nums[6] == nums[5] + 1 (does not exist, index 6 out of range). No next number, so the range ends at 'end' = 5.

💡 End of array means the last number is either a single or end of a range.

Line:

while end + 1 < n and nums[end + 1] == nums[end] + 1:
    end += 1

💡 No further consecutive numbers after 7, so this is a single number range.

insert

Record the last single number '7'

Since 'start' == 'end', append the single number '7' as a string to the 'ranges' list.

💡 Single numbers are recorded as themselves without arrows.

Line:

if start == end:
    ranges.append(str(nums[start]))
else:
    ranges.append(f"{nums[start]}->{nums[end]}")

💡 Single elements are represented simply as strings.

shrink

Move 'start' pointer beyond array length to end loop

Set 'start' to 'end + 1' (6), which is equal to the length of the array, ending the while loop.

💡 Moving 'start' beyond the last index signals completion of processing.

Line:

start = end + 1

💡 The algorithm has processed all elements and built the full summary ranges list.

reconstruct

Return the final list of summary ranges

Return the 'ranges' list containing all recorded ranges: ['0->2', '4->5', '7'].

💡 Returning the final result completes the algorithm.

Line:

return ranges

💡 The final output matches the expected summary ranges for the input array.

from typing import List

def summaryRanges(nums: List[int]) -> List[str]:
    ranges = []  # STEP 1
    n = len(nums)  # STEP 1
    start = 0  # STEP 1
    while start < n:  # STEP 16
        end = start  # STEP 2,8,13
        while end + 1 < n and nums[end + 1] == nums[end] + 1:  # STEP 3,4,9,10,14
            end += 1  # STEP 3,4,9
        if start == end:  # STEP 6,15
            ranges.append(str(nums[start]))  # STEP 6,15
        else:
            ranges.append(f"{nums[start]}->{nums[end]}")  # STEP 6,11
        start = end + 1  # STEP 7,12,16
    return ranges  # STEP 17

if __name__ == "__main__":
    print(summaryRanges([0,1,2,4,5,7]))  # ['0->2', '4->5', '7']

📊

Summary Ranges - Watch the Algorithm Execute, Step by Step

Watching each pointer move and each range decision helps you understand how to track sequences efficiently without extra space or complex data structures.

insert

empty map

Result: []

insert

empty map

Result: []

check

Lookup: 1→ true

empty map

Result: []

check

Lookup: 2→ true

empty map

Result: []

check

Lookup: 3→ false

empty map

Result: []

insert

0->2→0

Result: ["0->2"]

update

0->2→0

Result: ["0->2"]

insert

0->2→0

Result: ["0->2"]

check

Lookup: 4→ true

0->2→0

Result: ["0->2"]

check

Lookup: 5→ false

0->2→0

Result: ["0->2"]

insert

0->2→0

4->5→1

Result: ["0->2","4->5"]

update

0->2→0

4->5→1

Result: ["0->2","4->5"]

insert

0->2→0

4->5→1

Result: ["0->2","4->5"]

check

Lookup: 6→ false

0->2→0

4->5→1

Result: ["0->2","4->5"]

insert

0->2→0

4->5→1

7→2

Result: ["0->2","4->5","7"]

update

0->2→0

4->5→1

7→2

Result: ["0->2","4->5","7"]

check

0->2→0

4->5→1

7→2

Result: ["0->2","4->5","7"]

Key Takeaways

✓ Two pointers efficiently identify consecutive sequences without extra space.

This insight is hard to see from code alone because the pointer movements and their role in tracking ranges are implicit.

✓ Ranges are recorded only when a break in consecutiveness is detected, minimizing operations.

Visually seeing when and why ranges are appended clarifies the algorithm's efficiency.

✓ Single numbers are handled as a special case, recorded without arrows.

The trace shows exactly when the algorithm decides a range is a single number, which is subtle in code.

Practice

1. Given the following code snippet, what is the returned value when the input string is "abc"?

easy

A. 2

B. 1

C. 3

D. 0

Solution

1. Step 1: Trace iterations for input "abc"

   Right=0: 'a', distinct_count=1, window=[0,0], max_length=1 Right=1: 'b', distinct_count=2, window=[0,1], max_length=2 Right=2: 'c', distinct_count=3 > 2 triggers while loop: Shrink from left: freq['a']--, distinct_count=2, left=1 Window=[1,2], max_length=max(2,2)=2 But max_length updated after while loop: max_length=max(2,2)=2 Actually, max_length is updated each iteration, so max_length=2 after right=1, then after right=2 window is [1,2] length=2

2. Step 2: Check final max_length

   Max length found is 2, but wait, initial max_length was 0, updated to 1, then 2, then 2 again. But the substring "bc" length=2, so final answer is 2.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Longest substring with at most 2 distinct chars in "abc" is "ab" or "bc" length 2 [OK]

Hint: Check window size after shrinking when distinct > 2 [OK]

Common Mistakes:

• Returning length as right - left instead of right - left + 1
• Not updating distinct_count correctly
• Off-by-one errors in window size

2. The following code attempts to check if an array can be paired such that sums are divisible by k. Which line contains a subtle bug that causes incorrect results on some inputs?

def canArrange(nums, k):
    if len(nums) % 2 != 0:
        return True
    freq = {}
    for num in nums:
        r = ((num % k) + k) % k
        freq[r] = freq.get(r, 0) + 1
    if freq.get(0, 0) % 2 != 0:
        return False
    for r in freq:
        if r != 0:
            complement = (k - r) % k
            if freq.get(r, 0) != freq.get(complement, 0):
                return False
    if k % 2 == 0 and freq.get(k // 2, 0) % 2 != 0:
        return False
    return True

medium

A. Line 5: computing remainder without adjusting for negative numbers

B. Line 2: returning True when array length is odd

C. Line 8: checking if freq[0] count is even

D. Line 13: checking if freq[k//2] count is even when k is even

Solution

1. Step 1: Check array length condition

   The code returns True if the array length is odd, but pairing requires even length. This causes incorrect True on odd-length arrays.

2. Step 2: Verify other lines

   Remainder calculation correctly adjusts for negative numbers. Other checks for freq[0] and freq[k//2] are correct.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Odd length arrays cannot be fully paired, so returning True is a bug [OK]

Hint: Array length must be even to form pairs [OK]

Common Mistakes:

• Returning True on odd length arrays
• Not adjusting modulo for negative numbers
• Ignoring freq[0] or freq[k/2] parity checks

3. Identify the bug in the following code snippet for the Minimum Operations to Reduce X to Zero problem:

medium

A. Line with 'if target == 0: return -1' should return len(nums) instead.

B. The while loop condition should be 'current_sum >= target' instead of '>'.

C. The initialization of max_len should be 0 instead of -1.

D. The final return should be 'max_len' instead of 'len(nums) - max_len'.

Solution

1. Step 1: Understand the edge case when target == 0

   If target == 0, it means totalSum == x, so removing all elements is needed, so minimum operations = length of array.

2. Step 2: Identify incorrect return value

   The code returns -1 incorrectly, which signals no solution, but the correct answer is to remove all elements.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Returning -1 here causes wrong output on full removal case [OK]

Hint: Check edge cases where x equals total sum [OK]

Common Mistakes:

• Returning -1 instead of full length when target == 0
• Misinterpreting no solution vs full removal
• Ignoring edge cases in initial checks

4. What is the time complexity of the optimal prefix sum with hash map solution for the Subarray Sum Equals K problem, given an input array of length n?

medium

A. O(n) because each element is processed once and hash map operations are O(1) on average

B. O(n) for iteration plus O(n) for prefix sum storage, total O(n) space and time

C. O(n log n) due to sorting or balanced tree operations in the hash map

D. O(n^2) due to nested loops over the array

Solution

1. Step 1: Identify main operations

   The algorithm iterates over the array once, updating running_sum and hash map counts.

2. Step 2: Analyze hash map operations

   Hash map insertions and lookups are O(1) on average, so total time is O(n).

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   No nested loops or sorting involved, so O(n) time complexity. [OK]

Hint: Hash map lookups are average O(1), not O(log n) [OK]

Common Mistakes:

• Confusing with brute force O(n²)
• Assuming hash map operations are O(log n)
• Counting prefix sum storage as extra time

5. Suppose the problem is modified so that elements can be negative and the bounds left and right can also be negative. Which of the following changes is necessary to correctly count the number of subarrays with bounded maximum in this scenario?

hard

A. Adjust the logic to reset length when encountering elements outside the [left, right] range, not just greater than right.

B. Modify countAtMost to track minimum instead of maximum because negative numbers invert ordering.

C. Use a segment tree or balanced BST to query maximums efficiently due to negative values.

D. No change needed; the existing approach works for negative numbers as well.

Solution

1. Step 1: Understand impact of negative numbers

   Negative numbers do not invalidate the approach, but the reset condition must consider elements outside the bounds, including those less than left.

2. Step 2: Adjust reset logic

   Reset 'length' when elements are either > right or < left, because subarrays with max outside [left, right] are invalid.

3. Step 3: Why other options fail

   Tracking minimum is incorrect; segment trees add unnecessary complexity; existing approach assumes only upper bound resets.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Reset on elements outside [left, right] range for correctness [OK]

Hint: Reset on elements outside both bounds, not just upper bound [OK]

Common Mistakes:

• Assuming no change needed
• Switching to minimum tracking
• Adding complex data structures unnecessarily