<p>Design a stack-like data structure to push elements to the stack and pop the most frequent element from the stack.</p> <p>Implement the <code>FreqStack</code> class:</p> <ul> <li><code>FreqStack()</code> constructs an empty frequency stack.</li> <li><code>void push(int val)</code> pushes an integer <code>val</code> onto the top of the stack.</li> <li><code>int pop()</code> removes and returns the most frequent element in the stack. <ul> <li>If there is a tie for the most frequent element, the element closest to the stack's top is removed and returned.</li> </ul> </li> </ul> <p> </p> <p><strong>Example 1:</strong></p> <pre> <strong>Input</strong> ["FreqStack", "push", "push", "push", "push", "push", "push", "pop", "pop", "pop", "pop"] [[], [5], [7], [5], [7], [4], [5], [], [], [], []] <strong>Output</strong> [null, null, null, null, null, null, null, 5, 7, 5, 4] <strong>Explanation</strong> FreqStack freqStack = new FreqStack(); freqStack.push(5); // The stack is [5] freqStack.push(7); // The stack is [5,7] freqStack.push(5); // The stack is [5,7,5] freqStack.push(7); // The stack is [5,7,5,7] freqStack.push(4); // The stack is [5,7,5,7,4] freqStack.push(5); // The stack is [5,7,5,7,4,5] freqStack.pop(); // return 5, as 5 is the most frequent. The stack becomes [5,7,5,7,4]. freqStack.pop(); // return 7, as 5 and 7 is the most frequent, but 7 is closest to the top. The stack becomes [5,7,5,4]. freqStack.pop(); // return 5, as 5 is the most frequent. The stack becomes [5,7,4]. freqStack.pop(); // return 4, as 4, 5 and 7 is the most frequent, but 4 is closest to the top. The stack becomes [5,7]. </pre> <p> </p> <p><strong>Constraints:</strong></p> <ul> <li><code>0 <= val <= 10<sup>9</sup></code></li> <li>At most <code>2 * 10<sup>4</sup></code> calls will be made to <code>push</code> and <code>pop</code>.</li> <li>It is guaranteed that there will be at least one element in the stack before calling <code>pop</code>.</li> </ul>