 Solution

Divide-and-conquer algorithms suit problems with largely independent, balanced subproblems and an inexpensive combination step. In merge sort, the two halves are independently sorted and merged in linear time. Its recurrence relation is $T(n)=T(\lfloor n/2\rfloor )+T(\lceil n/2\rceil )+\Theta (n)$, giving $\Theta (n\log n)$ time complexity.

Dynamic programming suits overlapping subproblems with a compact state description. For the coin change problem, let $C(v)$ be the minimum number of coins needed to make amount $v$. With positive denominations $D$ containing $1$,Every optimal nonempty answer has a last coin $d$ and an optimal answer for $v-d$; otherwise replacing the remainder improves it. Filling $C(0),\dots ,C(V)$ takes $O(V|D|)$ time complexity and $O(V)$ space complexity. Repeated amounts are computed just once. This bound is in the numeric amount $V$, not its binary input length.

A greedy algorithm suits problems where one can prove a greedy-choice property. For unweighted interval scheduling, choose an interval with earliest finishing time, discard overlapping intervals, and repeat. Among optimal schedules, replace the first interval by the earliest-finishing one. The replacement finishes no later, so every later interval remains feasible. Applying the same argument to the remaining intervals proves optimality. After sorting by finish time, a single scan suffices.

These examples separate the relevant structural reasons: independent halves, repeated amount states, and an exchange argument certifying each irreversible choice.