Number of Subsequences That Satisfy the Given Sum Condition
A medium-tier problem at 44% community acceptance, tagged with Array, Two Pointers, Binary Search. Reported in interviews at Turing and 0 others.
You're staring at an array, and the problem asks you to count subsequences where the first element plus the last element equals a target sum. This one trips people up because they think they need to enumerate subsequences, which spirals into exponential complexity. The real edge is recognizing that once you fix which elements are first and last, the middle elements are free to pick or skip. Turing has reportedly asked this. It's medium-difficulty by rating but the pattern isn't intuitive, which is exactly why you need a safety net. If you hit this live in your OA and freeze on the counting trick, StealthCoder surfaces a working solution in seconds, invisible to the proctor.
Companies that ask "Number of Subsequences That Satisfy the Given Sum Condition"
Number of Subsequences That Satisfy the Given Sum Condition 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 to sort the array, then use two pointers: one at the start, one at the end. For each valid pair where arr[left] + arr[right] equals the target, every element between them can be included or excluded in the subsequence, giving you 2^(right - left - 1) valid subsequences. The pitfall is trying to iterate through all subsequences explicitly or using DFS without seeing the combinatorial shortcut. Sorting removes duplicates and lets you skip impossible pairs fast. Two pointers means you slide inward, shrinking the window until the sum matches. Binary search is an alternative when you need to find positions but the two-pointer greedy march is cleaner. If you haven't drilled this pattern or the math of counting combinations from a fixed span, StealthCoder is your hedge for the live assessment.
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Number of Subsequences That Satisfy the Given Sum Condition recycles across companies for a reason. It's medium-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.
Number of Subsequences That Satisfy the Given Sum Condition interview FAQ
Why does sorting help here when subsequences don't care about order?+
Sorting lets you use two pointers to prune impossible pairs in O(n log n) time. Subsequences preserve relative order from the original array, but once you identify valid first-last pairs, the count of middle elements is fixed. You're not permuting, you're counting combinations.
What's the difference between a subsequence and a subarray in this problem?+
A subarray must be contiguous. A subsequence can skip elements but must preserve relative order from the original. This problem counts subsequences, so once you pick a first and last element (even if far apart), every element between them in the sorted array can be independently included or skipped.
Is this problem still asked at major tech companies?+
Reports show it's in circulation at companies like Turing. It's medium-difficulty but the combinatorial counting trick is not obvious, so it surfaces to catch people who can code but don't recognize the pattern.
How do you handle duplicates in the array?+
Sorting groups duplicates together. When left and right point to equal values, all elements between them (including duplicates) contribute to the power-of-two count. If left == right, there's exactly one element, giving 2^0 = 1 subsequence.
What if no pairs sum to the target?+
The two-pointer loop will converge without finding matches, and you'll return 0. If arr[left] + arr[right] is less than target, increment left. If greater, decrement right. If equal, count the 2^(right - left - 1) and move inward.
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