Palantir·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Jun 2026

Summary

Palantir SWE coding round, one problem the whole session. The question was algorithmic and a bit unusual, not your standard leetcode fare.

Questions Asked (1)

You're given a binary array. Each operation lets you cycle a row's elements left or right. What's the minimum number of operations to produce a column made entirely of 1s? Return 0 if it's impossible.

Algorithms & Data Structures

Author's notes

Took me a minute to even figure out what 'possible' means here.

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AI HintsAI Generated

Suggested Approach

For each column, determine the minimum rotations needed to make all rows have a 1 in that column, then take the minimum across columns. If any column has no 1 in a row, that column is impossible; if all columns are impossible, return 0.

Pro tip: Clarify that 'cycle a row' means shifting the entire row left or right by one position per operation, and that operations on different rows are independent. Also, mention that you can precompute the positions of 1s in each row to quickly check column feasibility.

1. Understand the problem and constraints

Confirm that each operation shifts a single row by one position (left or right) and that you can perform operations on multiple rows. The goal is to align a column of 1s with minimum total shifts.

2. Preprocess row data

For each row, record the indices where 1s appear. This allows O(1) lookup of whether a row can contribute a 1 to a given column and the required shift.

3. Evaluate each column

For each column index c, check every row: if the row has a 1, compute the minimal shift (left or right) to move that 1 to column c; if no 1, the column is impossible. Sum the minimal shifts for all rows to get the cost for that column.

4. Find the minimum cost

Track the minimum total shifts across all feasible columns. If no column is feasible, return 0 as specified.

5. Analyze complexity and edge cases

The algorithm runs in O(n * m) time where n is rows and m is columns, with O(n * m) space for storing positions. Discuss edge cases like empty array, all zeros, or already aligned columns.

Key Points to Mention

Independence of row operations: shifting one row does not affect others.
Minimal shift calculation: for a 1 at index j in a row of length m, the shift to column c is min(|j-c|, m - |j-c|).
Feasibility check: a column is only possible if every row has at least one 1.
Time and space complexity: O(n*m) time and O(n*m) space, which is optimal for this problem.
Edge cases: return 0 if no column can be formed; handle empty input gracefully.
Optimization: precompute positions of 1s per row to avoid scanning the entire row for each column.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.