New problems done, including april day 10 challenge

problems/155/main.py

+class MinStack:
+
+    def __init__(self):
+        self.stack = []
+        self.minEl = None
+
+    def push(self, x: int) -> None:
+        node = Node(x)
+        if self.minEl is None or node.value <= self.minEl.value:
+            node.lastMin = self.minEl
+            self.minEl = node
+        self.stack.append(node)
+
+    def pop(self) -> None:
+        if len(self.stack) > 0:
+            ret = self.stack[len(self.stack) - 1]
+            if ret.value == self.minEl.value:
+                self.minEl = self.minEl.lastMin
+            self.stack = self.stack[0:len(self.stack) - 1]
+            return ret.value
+        return None
+
+    def top(self) -> int:
+        if len(self.stack) > 0:
+            return self.stack[len(self.stack) - 1].value
+        return None
+
+    def getMin(self) -> int:
+        return self.minEl.value
+
+    def __str__(self):
+        str_result = ""
+        for x in self.stack:
+            str_result += "\n"
+            str_result += str(x)
+        return str_result
+    
+class Node:
+    def __init__(self, x, lastMin = None):
+        self.value = x
+        self.lastMin = lastMin
+        
+    def __str__(self):
+        return f"Value: {str(self.value)} Last Min: {str(self.lastMin)}"
+
+# Your MinStack object will be instantiated and called as such:
+values = []
+obj = MinStack()
+obj.push(-2)
+obj.push(0)
+obj.push(-3)
+values.append(obj.getMin())
+values.append(obj.pop())
+values.append(obj.top())
+values.append(obj.getMin())
+
+print("Expected: [-3, -3, 0, -2]")
+print("Got:", values)

problems/155/problem.txt

+Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
+
+    push(x) -- Push element x onto stack.
+    pop() -- Removes the element on top of the stack.
+    top() -- Get the top element.
+    getMin() -- Retrieve the minimum element in the stack.
+
+ 
+
+Example:
+
+MinStack minStack = new MinStack();
+minStack.push(-2);
+minStack.push(0);
+minStack.push(-3);
+minStack.getMin();   --> Returns -3.
+minStack.pop();
+minStack.top();      --> Returns 0.
+minStack.getMin();   --> Returns -2.
+

problems/155/run.sh

+#!/bin/bash
+
+python3 main.py

problems/26/main.c

+#include <stdio.h>
+
+int removeDuplicates(int*, int);
+int removeDuplicates(int* nums, int numsSize) {
+    if (numsSize == 0) {
+        return 0;
+    }
+    int shift = 0;
+    int last = nums[0] + 1; // This way the first item will never be a duplicate
+    for (int i = 0; i < numsSize; i++) {
+        int value = nums[i];
+        nums[i - shift] = value;
+        if (value == last) {
+            shift++;
+        }
+        last = value;
+    }
+    return numsSize - shift;
+}
+
+int main() {
+    printf("Expetected: Length 5, [0, 1, 2, 3, 4]\n");
+    int arr[10] = {0,0,1,1,1,2,2,3,3,4};
+    int newLength = removeDuplicates(&arr[0], 10); // &arr[0] is a hack to stop warning about static sized pointer being passed
+    printf("Got: Length: %d,", newLength);
+    for (int i = 0; i < newLength; i++) {
+        printf(" %d ", arr[i]);
+    }
+    printf("\n");
+    return 0;
+}

problems/26/problem.txt

+Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length.
+
+Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
+
+Example 1:
+
+Given nums = [1,1,2],
+
+Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.
+
+It doesn't matter what you leave beyond the returned length.
+
+Example 2:
+
+Given nums = [0,0,1,1,1,2,2,3,3,4],
+
+Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.
+
+It doesn't matter what values are set beyond the returned length.
+
+Clarification:
+
+Confused why the returned value is an integer but your answer is an array?
+
+Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.
+
+Internally you can think of this:
+
+// nums is passed in by reference. (i.e., without making a copy)
+int len = removeDuplicates(nums);
+
+// any modification to nums in your function would be known by the caller.
+// using the length returned by your function, it prints the first len elements.
+for (int i = 0; i < len; i++) {
+    print(nums[i]);
+}

problems/26/run.sh

+#!/bin/bash
+
+gcc -o main main.c
+./main
+rm main