K Inverse Pairs Array
A hard-tier problem at 49% community acceptance, tagged with Dynamic Programming. Reported in interviews at Works Applications and 0 others.
K Inverse Pairs Array is a hard DP problem that shows up in live assessments with brutal regularity. If you're interviewing at Works Applications or any company that loves dynamic programming, this one's on the table. The problem asks you to count arrays of length n containing elements from 1 to n where exactly k pairs (i, j) satisfy i < j and arr[i] > arr[j]. It's not a brute-force problem. The naive enumeration explodes combinatorially. Most candidates freeze on the state definition. If you hit this during a live OA and the pattern doesn't click, StealthCoder surfaces the DP transition in seconds, invisible to the proctor.
Companies that ask "K Inverse Pairs Array"
K Inverse Pairs Array is the kind of problem that decides whether you pass. StealthCoder reads the problem on screen and surfaces a working solution in under 2 seconds. Invisible to screen share. The proctor sees nothing. Made by a working Amazon engineer who got tired of watching qualified friends bomb OAs they'd solve cold in an IDE.
Get StealthCoderThe trick is redefining the problem in DP terms. Instead of thinking about permutations and counting inversions after the fact, build the answer element by element. The key insight: when you insert the next number into a partially built array, you control exactly how many new inverse pairs it creates based on its position. That position depends on how many elements to its left are greater. You need a 2D state: dp[i][j] = count of arrays of length i with exactly j inverse pairs. The transition requires you to think about where the new element lands among the i previously placed elements. Most candidates miss that you can calculate the contribution directly without iterating through all permutations. The recurrence is dense and the optimization from O(n^2 * k^2) to O(n * k^2) using prefix sums separates solvers from blockers. If the math stalls you mid-interview, StealthCoder gives you the working solution and transition logic without explanation overhead.
Pattern tags
You know the problem.
Make sure you actually pass it.
K Inverse Pairs Array recycles across companies for a reason. It's hard-tier, and most candidates blank under the timer. StealthCoder is the hedge: an AI overlay invisible during screen share. It reads the problem and surfaces a working solution in under 2 seconds. Made by a working Amazon engineer who got tired of watching qualified friends bomb OAs they'd solve cold in an IDE. Works on HackerRank, CodeSignal, CoderPad, and Karat.
K Inverse Pairs Array interview FAQ
Is K Inverse Pairs Array really asked at major companies?+
It appears in live assessments, though rarely at household-name scale. Works Applications is the documented company. It's a hard DP problem that filters out candidates who can't move past brute-force thinking. Not FAANG-frequent, but it shows up when companies want to test pure algorithmic depth.
What's the acceptance rate and why is it so low?+
The acceptance rate is roughly 49 percent, which means nearly half fail. Most submissions are either wrong logic (treating it as a permutation-counting problem) or timeout from O(n^3) or worse approaches. The DP state and transition are non-obvious, and the jump to an optimized solution requires solid understanding of how DP interacts with combinatorics.
Is this problem just about counting permutations?+
No. You're not enumerating permutations and counting inversions in each. You're building arrays incrementally and calculating inverse pair contributions on the fly. The insight that inserting element x at position p in an array of i elements creates exactly a specific number of new inversions is what unlocks the DP. Permutation thinking leads to timeout.
How does the DP transition work at a high level?+
You maintain dp[i][j] = ways to form an array of i elements with j inverse pairs. When you place the (i+1)-th element, its position determines how many inversions it adds. If it lands at position p among i existing elements, it creates p new inversions. You sum contributions across all valid positions and previous inverse pair counts.
What's the time-complexity trap?+
The naive O(n^2 * k^2) loop over positions and previous states times out on large n and k. The fix is prefix-sum optimization to collapse the innermost loop. If you submit without this optimization, expect TLE even on medium inputs. The optimization is rarely obvious during a live assessment under pressure.
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