Find ID of Soldier

There are N soldiers standing in a line, with IDs from 1 to N, in ascending order. They are participating in an exercise consisting of Q actions. During the ith action, the Major calls S numbers rowi and coli. The soldiers at the rowith and colith positions swap places; then the soldiers at (rowi+1)th and (coli-1)th positions swap places, and so on until (rowi+m) < (coli-m). Each of the soldier's IDs will be covered in the range [rowi, coli] for at most one action.
Write an algorithm to find the ID of the soldier at Kth position in the line after all the actions are completed.
Input
The first line of the input consists of an integer- num, representing the number of soldiers (N).
The second line consists of two space-separated integers- actions and numSoldiers, representing the number of actions (Q) and number of soldiers called by the Major (S), respectively.
The next Q lines consist of S space-separated integers - rowi and coli, representing the positions of the soldiers initially called for the ith action.
The last line consists of an integer- posSoldier, representing the position of the soldier whose ID is requested to be found after Q actions (K).
Output
Print an integer representing the ID of the Kth position soldier in the line after Q actions.

Function
findIdOfSoldier(num: int, actions: int, numSoldiers: int, swaps: int[][], posSoldier: int) → int

Examples
Example 1
num = 10
actions = 2
numSoldiers = 2
swaps = [[1, 5], [6, 10]]
posSoldier = 1
return = 5
Step1: After the 1st action, the position of soldiers is in the order: 5 4 3 2 1 6 7 8 9 10.
Step2: After the 2nd action, the position of soldiers is in the order: 5 4 3 2 1 10 9 8 7 6.
Step3: The ID of the soldier at position 1 is 5.
So, the output is 5.

Constraints
1 <= posSoldier <= num <= 10^9
1 <= actions <= 10^5
1 <= rowi <= coli <= num
1 <= i <= actions

Each action reverses the segment [row, col]. The statement guarantees every soldier is covered by at most one action, so the ranges never overlap. That's the whole trick. Position K is either outside every range, so the ID is just K, or inside exactly one range [l, r]. Reversing maps position K to l + r - K, and that's the answer. Scan the Q actions once, find the range containing K, and return l + r - K. If none contains it, return K. That's O(Q) time and O(1) extra space. The pitfall is simulating swaps or allocating an array, which fails at 10^9. Another trap is applying the reflection more than once, which is wrong because the ranges are disjoint. Read input carefully, since the S parameter is always 2 in practice. If you freeze under the clock, StealthCoder is the hedge that surfaces this reflection formula during the live OA.

Memorize the pattern. If you can't, run StealthCoder. The proctor sees the IDE. They don't see what's behind it.