We create a merged list containing all elements from the input lists, each paired with the index of its originating list. This prepares for sorting and sliding window operations.
💡 Merging all elements into one list with their source indices allows us to handle all values uniformly and track coverage of each list.
Line:for i, lst in enumerate(nums):
for val in lst:
merged.append((val, i))
💡 This step sets up the data structure needed to apply the two-pointer sliding window technique across all lists simultaneously.
setup
Sort the merged list by element value
We sort the merged list by the element values to enable the sliding window to move over a sorted sequence, which is essential for finding the smallest range.
💡 Sorting ensures the window moves over elements in ascending order, making it easier to find minimal ranges.
Line:merged.sort(key=lambda x: x[0])
💡 Sorted order allows the two pointers to efficiently explore candidate ranges without missing smaller ones.
setup
Initialize pointers and counters
We initialize the left pointer at index 0, the right pointer will start moving from 0, and counters to track how many lists are covered in the current window.
💡 Setting up pointers and counters is crucial for managing the sliding window and coverage tracking.
Line:left = 0
count = Counter()
have = 0
need = k
min_range = float('inf')
res = []
💡 These variables control the window boundaries and track coverage to find valid ranges.
expand
Expand right pointer to include (0,1)
We move the right pointer to index 0, include element 0 from list 1, and update coverage count. Since this is the first element from list 1, coverage increases.
💡 Expanding right pointer adds elements to the window, increasing coverage of lists.
Line:val, idx = merged[right]
count[idx] += 1
if count[idx] == 1:
have += 1
💡 Including a new list's element increases 'have', moving closer to covering all lists.
expand
Expand right pointer to include (4,0)
Right pointer moves to index 1, adding element 4 from list 0. This is the first element from list 0, so coverage increases again.
💡 Each new list element included increases coverage until all lists are covered.
Line:val, idx = merged[right]
count[idx] += 1
if count[idx] == 1:
have += 1
💡 Coverage now includes lists 0 and 1, still missing list 2.
expand
Expand right pointer to include (5,2)
Right pointer moves to index 2, adding element 5 from list 2. This completes coverage of all three lists, so 'have' equals 'need'.
💡 Once all lists are covered, we can try to shrink the window to find smaller ranges.
Line:val, idx = merged[right]
count[idx] += 1
if count[idx] == 1:
have += 1
💡 The window now covers all lists, enabling range evaluation and potential shrinking.
shrink
Evaluate and shrink window from left
Since all lists are covered, we check if the current window [0,2] is smaller than the previous best. It is, so we update the smallest range to [0,5]. Then we shrink from the left by removing element (0,1).
💡 Shrinking the window tries to find a smaller valid range while maintaining coverage.
Line:while have == need:
left_val, left_idx = merged[left]
if val - left_val < min_range:
min_range = val - left_val
res = [left_val, val]
count[left_idx] -= 1
if count[left_idx] == 0:
have -= 1
left += 1
💡 Shrinking from left can reduce range size but risks losing coverage, so we track coverage carefully.
expand
Expand right pointer to include (9,1)
Right pointer moves to index 3, adding element 9 from list 1. Coverage for list 1 increases again, but since it was already covered, 'have' remains the same.
💡 Expanding right pointer continues to explore new ranges and possibly regain coverage lost by shrinking.
Line:val, idx = merged[right]
count[idx] += 1
if count[idx] == 1:
have += 1
💡 Adding elements from already covered lists keeps coverage stable but may help find smaller ranges.
shrink
Shrink window from left to [4,9]
We shrink the window by moving left pointer from index 1 to 2, removing element (4,0). Coverage for list 0 decreases but remains covered because count is still positive.
💡 Shrinking tries to minimize the range while maintaining coverage.
Line:count[left_idx] -= 1
if count[left_idx] == 0:
have -= 1
left += 1
💡 Coverage counts allow safe shrinking without losing coverage prematurely.
expand
Expand right pointer to include (10,0)
Right pointer moves to index 4, adding element 10 from list 0. Coverage for list 0 increases again, maintaining full coverage.
💡 Expanding right pointer explores new elements to find potentially smaller ranges.
Line:val, idx = merged[right]
count[idx] += 1
if count[idx] == 1:
have += 1
💡 Coverage remains full, allowing further shrinking attempts.
shrink
Shrink window from left to [5,10]
We shrink the window by moving left pointer from index 2 to 3, removing element (5,2). Coverage for list 2 decreases but remains covered because count is still positive.
💡 Shrinking continues to try to minimize the range while keeping coverage intact.
Line:count[left_idx] -= 1
if count[left_idx] == 0:
have -= 1
left += 1
💡 Coverage counts allow safe shrinking without losing coverage prematurely.
expand
Expand right pointer to include (12,1)
Right pointer moves to index 5, adding element 12 from list 1. Coverage for list 1 increases again, maintaining full coverage.
💡 Expanding right pointer explores new elements to find potentially smaller ranges.
Line:val, idx = merged[right]
count[idx] += 1
if count[idx] == 1:
have += 1
💡 Coverage remains full, allowing further shrinking attempts.
expand
Expand right pointer to include (15,0)
Right pointer moves to index 6, adding element 15 from list 0. Coverage for list 0 increases again, maintaining full coverage.
💡 Expanding right pointer explores new elements to find potentially smaller ranges.
Line:val, idx = merged[right]
count[idx] += 1
if count[idx] == 1:
have += 1
💡 Coverage remains full, allowing further shrinking attempts.
expand
Expand right pointer to include (18,2)
Right pointer moves to index 7, adding element 18 from list 2. Coverage for list 2 increases again, maintaining full coverage.
💡 Expanding right pointer explores new elements to find potentially smaller ranges.
Line:val, idx = merged[right]
count[idx] += 1
if count[idx] == 1:
have += 1
💡 Coverage remains full, allowing further shrinking attempts.
expand
Expand right pointer to include (20,1)
Right pointer moves to index 8, adding element 20 from list 1. Coverage remains full. We now check if the current window [3,8] gives a smaller range than before.
💡 When coverage is full, we evaluate the current window's range and update the smallest range if better.
Line:val, idx = merged[right]
count[idx] += 1
if count[idx] == 1:
have += 1
while have == need:
left_val, left_idx = merged[left]
if val - left_val < min_range or (val - left_val == min_range and left_val < (res[0] if res else float('inf'))):
min_range = val - left_val
res = [left_val, val]
💡 We found a smaller valid range covering all lists, so we update the result.
shrink
Shrink window from left to [10,20]
We shrink the window by moving left pointer from index 3 to 4, removing element (9,1). Coverage remains full, so we update the smallest range to [10,20].
💡 Shrinking the window while coverage remains full helps find the minimal range.
Line:count[left_idx] -= 1
if count[left_idx] == 0:
have -= 1
left += 1
if val - left_val < min_range or (val - left_val == min_range and left_val < (res[0] if res else float('inf'))):
min_range = val - left_val
res = [left_val, val]
💡 Shrinking can improve the range while maintaining coverage.
expand
Expand right pointer to include (22,2)
Right pointer moves to index 9, adding element 22 from list 2. Coverage remains full, so we check if the new window [4,9] gives a smaller range.
💡 Expanding right pointer explores new elements to find potentially smaller ranges.
Line:val, idx = merged[right]
count[idx] += 1
if count[idx] == 1:
have += 1
while have == need:
left_val, left_idx = merged[left]
if val - left_val < min_range or (val - left_val == min_range and left_val < (res[0] if res else float('inf'))):
min_range = val - left_val
res = [left_val, val]
💡 We are approaching the optimal smallest range covering all lists.
shrink
Shrink window to [20,24]
We shrink the window by moving left pointer from index 4 to 10 through several steps, removing elements until the window covers [20,24]. This is the smallest range found that covers all lists.
💡 Final shrinking finds the minimal window covering all lists.
Line:while have == need:
left_val, left_idx = merged[left]
if val - left_val < min_range or (val - left_val == min_range and left_val < (res[0] if res else float('inf'))):
min_range = val - left_val
res = [left_val, val]
count[left_idx] -= 1
if count[left_idx] == 0:
have -= 1
left += 1
💡 The smallest valid range is [20,24], covering all lists with minimal width.
reconstruct
Final answer found
The algorithm finishes with the smallest range [20,24] that covers all lists. This range includes elements 24 (list 0), 20 (list 1), and 22 (list 2).
💡 The final range is the minimal window covering all lists, found by expanding and shrinking the sliding window.
Line:return res
💡 The two-pointer sliding window approach efficiently finds the smallest range covering all lists.
from collections import Counter
def smallestRangeTwoPointers(nums):
merged = []
k = len(nums)
for i, lst in enumerate(nums): # STEP 1
for val in lst:
merged.append((val, i))
merged.sort(key=lambda x: x[0]) # STEP 2
count = Counter()
have = 0
need = k
left = 0 # STEP 3
min_range = float('inf')
res = []
for right in range(len(merged)):
val, idx = merged[right] # STEP 4+ expanding right
count[idx] += 1
if count[idx] == 1:
have += 1
while have == need: # STEP 7+ shrinking left
left_val, left_idx = merged[left]
if val - left_val < min_range or (val - left_val == min_range and left_val < (res[0] if res else float('inf'))):
min_range = val - left_val
res = [left_val, val]
count[left_idx] -= 1
if count[left_idx] == 0:
have -= 1
left += 1
return res
if __name__ == '__main__':
nums = [[4,10,15,24,26],[0,9,12,20],[5,18,22,30]]
print(smallestRangeTwoPointers(nums))
📊
Smallest Range Covering K Lists - Watch the Algorithm Execute, Step by Step
Watching the algorithm step-by-step reveals how the sliding window expands and shrinks to cover all lists efficiently, which is hard to grasp from code alone.
Step 1/19
·Active fill★Answer cell
setup
(4,0)
0
(10,0)
1
(15,0)
2
(24,0)
3
(26,0)
4
(0,1)
5
(9,1)
6
(12,1)
7
(20,1)
8
(5,2)
9
(18,2)
10
(22,2)
11
(30,2)
12
Result: "[]"
setup
(0,1)
0
(4,0)
1
(5,2)
2
(9,1)
3
(10,0)
4
(12,1)
5
(15,0)
6
(18,2)
7
(20,1)
8
(22,2)
9
(24,0)
10
(26,0)
11
(30,2)
12
Result: "[]"
setup
left
(0,1)
0
(4,0)
1
(5,2)
2
(9,1)
3
(10,0)
4
(12,1)
5
(15,0)
6
(18,2)
7
(20,1)
8
(22,2)
9
(24,0)
10
(26,0)
11
(30,2)
12
Result: "[]"
expand
right
(0,1)
0
(4,0)
1
(5,2)
2
(9,1)
3
(10,0)
4
(12,1)
5
(15,0)
6
(18,2)
7
(20,1)
8
(22,2)
9
(24,0)
10
(26,0)
11
(30,2)
12
Result: "[]"
expand
left
(0,1)
0
right
(4,0)
1
(5,2)
2
(9,1)
3
(10,0)
4
(12,1)
5
(15,0)
6
(18,2)
7
(20,1)
8
(22,2)
9
(24,0)
10
(26,0)
11
(30,2)
12
Result: "[]"
expand
left
(0,1)
0
(4,0)
1
right
(5,2)
2
(9,1)
3
(10,0)
4
(12,1)
5
(15,0)
6
(18,2)
7
(20,1)
8
(22,2)
9
(24,0)
10
(26,0)
11
(30,2)
12
Result: "[]"
shrink
(0,1)
0
left
(4,0)
1
right
(5,2)
2
(9,1)
3
(10,0)
4
(12,1)
5
(15,0)
6
(18,2)
7
(20,1)
8
(22,2)
9
(24,0)
10
(26,0)
11
(30,2)
12
Result: "[0,5]"
expand
(0,1)
0
left
(4,0)
1
(5,2)
2
right
(9,1)
3
(10,0)
4
(12,1)
5
(15,0)
6
(18,2)
7
(20,1)
8
(22,2)
9
(24,0)
10
(26,0)
11
(30,2)
12
Result: "[0,5]"
shrink
(0,1)
0
(4,0)
1
left
(5,2)
2
right
(9,1)
3
(10,0)
4
(12,1)
5
(15,0)
6
(18,2)
7
(20,1)
8
(22,2)
9
(24,0)
10
(26,0)
11
(30,2)
12
Result: "[0,5]"
expand
(0,1)
0
(4,0)
1
left
(5,2)
2
(9,1)
3
right
(10,0)
4
(12,1)
5
(15,0)
6
(18,2)
7
(20,1)
8
(22,2)
9
(24,0)
10
(26,0)
11
(30,2)
12
Result: "[0,5]"
shrink
(0,1)
0
(4,0)
1
(5,2)
2
left
(9,1)
3
right
(10,0)
4
(12,1)
5
(15,0)
6
(18,2)
7
(20,1)
8
(22,2)
9
(24,0)
10
(26,0)
11
(30,2)
12
Result: "[0,5]"
expand
(0,1)
0
(4,0)
1
(5,2)
2
left
(9,1)
3
(10,0)
4
right
(12,1)
5
(15,0)
6
(18,2)
7
(20,1)
8
(22,2)
9
(24,0)
10
(26,0)
11
(30,2)
12
Result: "[0,5]"
expand
(0,1)
0
(4,0)
1
(5,2)
2
left
(9,1)
3
(10,0)
4
(12,1)
5
right
(15,0)
6
(18,2)
7
(20,1)
8
(22,2)
9
(24,0)
10
(26,0)
11
(30,2)
12
Result: "[0,5]"
expand
(0,1)
0
(4,0)
1
(5,2)
2
left
(9,1)
3
(10,0)
4
(12,1)
5
(15,0)
6
right
(18,2)
7
(20,1)
8
(22,2)
9
(24,0)
10
(26,0)
11
(30,2)
12
Result: "[0,5]"
expand
(0,1)
0
(4,0)
1
(5,2)
2
left
(9,1)
3
(10,0)
4
(12,1)
5
(15,0)
6
(18,2)
7
right
(20,1)
8
(22,2)
9
(24,0)
10
(26,0)
11
(30,2)
12
Result: "[9,20]"
shrink
(0,1)
0
(4,0)
1
(5,2)
2
(9,1)
3
left
(10,0)
4
(12,1)
5
(15,0)
6
(18,2)
7
right
(20,1)
8
(22,2)
9
(24,0)
10
(26,0)
11
(30,2)
12
Result: "[10,20]"
expand
(0,1)
0
(4,0)
1
(5,2)
2
(9,1)
3
left
(10,0)
4
(12,1)
5
(15,0)
6
(18,2)
7
(20,1)
8
right
(22,2)
9
(24,0)
10
(26,0)
11
(30,2)
12
Result: "[10,20]"
shrink
(0,1)
0
(4,0)
1
(5,2)
2
(9,1)
3
(10,0)
4
(12,1)
5
(15,0)
6
(18,2)
7
(20,1)
8
(22,2)
9
right
(24,0)
10
(26,0)
11
(30,2)
12
Result: "[20,24]"
record
(0,1)
0
(4,0)
1
(5,2)
2
(9,1)
3
(10,0)
4
(12,1)
5
(15,0)
6
(18,2)
7
(20,1)
8
(22,2)
9
(24,0)
10
(26,0)
11
(30,2)
12
Result: "[20,24]"
Key Takeaways
✓ Merging and sorting all elements from k lists allows a unified sliding window approach.
This insight is hard to see from code alone because it involves understanding the data transformation before the main algorithm.
✓ The two-pointer sliding window expands to cover all lists and shrinks to minimize the range.
Visualizing pointer movements clarifies how coverage and range size are balanced dynamically.
✓ Coverage counts track how many lists are included, controlling when to shrink or expand the window.
Seeing coverage counts change step-by-step reveals why the window moves as it does.
Practice
(1/5)
1. You need to place n queens on an nxn chessboard so that no two queens threaten each other. Which algorithmic approach guarantees finding all valid configurations efficiently by pruning invalid partial solutions early?
easy
A. Backtracking that tries all column positions per row and prunes invalid placements using sets or bitmasks
B. Greedy algorithm that places queens row by row choosing the first safe column
C. Dynamic programming that stores subproblems of partial queen placements by rows and columns
D. Divide and conquer by splitting the board into smaller sub-boards and solving independently
Solution
Step 1: Understand problem constraints
The problem requires placing queens so none attack each other, which involves checking columns and diagonals.
Step 2: Identify approach that prunes invalid states
Backtracking tries all placements but prunes invalid ones early using sets or bitmasks, guaranteeing all solutions without brute force.
Final Answer:
Option A -> Option A
Quick Check:
Backtracking with pruning is standard for N-Queens [OK]
Hint: Backtracking prunes invalid queen placements early [OK]
Common Mistakes:
Thinking greedy placement always works
Confusing DP with backtracking
Assuming independent subproblems in divide and conquer
2. Given the following code snippet implementing the two-pointer approach for trapping rain water, what is the value of the variable water after the first iteration of the while loop when the input is [2, 0, 2]?
easy
A. 0
B. 2
C. 1
D. 3
Solution
Step 1: Initialize variables
left=0, right=2, left_max=0, right_max=0, water=0
Step 2: First iteration of while loop
height[left]=2, height[right]=2, since height[left] is not less than height[right], else branch runs. height[right]=2 ≥ right_max=0, so right_max=2, right decrements to 1, water remains 0.
Final Answer:
Option A -> Option A
Quick Check:
Water not added on first iteration because max updated [OK]
Hint: Water added only when current height < max height on that side [OK]
Common Mistakes:
Adding water before max height is updated
Confusing left and right pointer conditions
Off-by-one errors in pointer movement
3. Consider the following buggy code snippet for merging two skylines. Which line contains the subtle bug that can cause incorrect skyline output by adding redundant points or missing height changes?
medium
A. Line where maxH is computed with max(h1, h2)
B. Line where merged appends [x, maxH] without checking last height
C. Line where i and j are incremented in the else block
D. Line where h1 and h2 are initialized to 0 at the start
Solution
Step 1: Identify the bug
The code appends every point without checking if the height changed, causing redundant points.
Step 2: Correct approach
Appending should happen only if merged is empty or last height differs from maxH to avoid duplicates.
Hint: Check if last height differs before appending new point [OK]
Common Mistakes:
Appending points even if height unchanged
Incorrectly incrementing indices causing skips
Not initializing heights properly
4. Suppose the Burst Balloons problem is modified so that you can burst the same balloon multiple times (balloons regenerate after bursting). Which of the following changes to the algorithm correctly adapts the solution to this variant?
hard
A. Modify the DP to consider infinite bursting by changing the recurrence to allow repeated indices and use memoization with state tracking of burst counts
B. Switch to a greedy approach that always bursts the balloon with the highest product value repeatedly
C. Use the same bottom-up DP but allow dp states to revisit the same balloon multiple times without changes
D. Use a top-down DFS without memoization to explore all sequences with repeated bursts
Solution
Step 1: Understand the new problem variant
Allowing multiple bursts of the same balloon means the state must track how many times each balloon has been burst or if it is still available.
Step 2: Adapt DP to handle repeated bursting
The original DP assumes each balloon bursts once. To handle repeats, the recurrence must be modified to track burst counts or states, and memoization must store these states to avoid recomputation.
Step 3: Evaluate options
Use the same bottom-up DP but allow dp states to revisit the same balloon multiple times without changes ignores state changes and is incorrect. Switch to a greedy approach that always bursts the balloon with the highest product value repeatedly is greedy and suboptimal. Use a top-down DFS without memoization to explore all sequences with repeated bursts is exponential without memoization. Modify the DP to consider infinite bursting by changing the recurrence to allow repeated indices and use memoization with state tracking correctly modifies DP with state tracking and memoization.
Final Answer:
Option A -> Option A
Quick Check:
State tracking and memoization needed for repeated bursts [OK]
Hint: Repeated bursting requires state tracking in DP [OK]
Common Mistakes:
Using original DP unchanged
Greedy approach
Ignoring memoization
5. Suppose the problem is modified so that each element in nums can be used multiple times for XOR queries (i.e., unlimited reuse), and queries remain the same. Which of the following changes to the optimized approach is correct to handle this variant efficiently?
hard
A. No change needed; the existing trie approach already supports unlimited reuse
B. Use a dynamic programming approach over XOR states instead of a trie
C. Modify the trie to store counts of inserted elements and adjust maxXor to consider reuse
D. Sort queries and nums but do not insert elements multiple times; reuse is implicit
Solution
Step 1: Understand reuse impact
Unlimited reuse means elements can be XORed multiple times, changing the problem semantics.
Step 2: Adjust trie to track counts
Trie nodes must store counts of inserted elements to allow multiple traversals representing reuse.
Step 3: Modify maxXor accordingly
maxXor must consider paths with repeated elements, enabled by counts, to find max XOR with reuse.
Final Answer:
Option C -> Option C
Quick Check:
Trie with counts supports reuse; naive trie does not [OK]
Hint: Reuse requires tracking counts in trie nodes [OK]
Common Mistakes:
Assuming no change needed
Switching to DP unnecessarily
Ignoring reuse in insertion logic
Recommended
Ready to prove it?
Pass the test and this pattern moves unknownstrongwhile earning +1 Grip toward your certificate.
