Merge pull request #2782 from Abe0770/main

Create 1675-minimize-deviation-in-array.cpp

Author: felivalencia3

Diff for: c/0024-swap-nodes-in-pairs.c

@@ -0,0 +1,41 @@
+/*
+  Given a linked list, swap every two adjacent nodes and return its head. 
+  You must solve the problem without modifying the values in the list's nodes (i.e., only nodes themselves may be changed.)
+
+  Ex. Input: head = [1,2,3,4]
+      Output: [2,1,4,3]
+
+  Time  : O(N)
+  Space : O(1)
+*/
+
+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     struct ListNode *next;
+ * };
+ */
+struct ListNode* swapPairs(struct ListNode* head) {
+    if (head == NULL || head->next == NULL) 
+        return head;
+
+    struct ListNode *new_head = head->next;
+    struct ListNode *prev = NULL;
+
+    while (head != NULL && head->next != NULL) {
+        struct ListNode *next_pair = head->next->next;
+        struct ListNode *second = head->next;
+
+        if (prev != NULL)
+            prev->next = second;
+
+        second->next = head;
+        head->next = next_pair;
+
+        prev = head;
+        head = next_pair;
+    }
+
+    return new_head;
+}

Diff for: cpp/1675-minimize-deviation-in-array.cpp

@@ -0,0 +1,43 @@
+/*
+  You are given an array nums of n positive integers.
+
+  You can perform two types of operations on any element of the array any number of times:
+
+    If the element is even, divide it by 2.
+        For example, if the array is [1,2,3,4], then you can do this operation on the last element, and the array will be [1,2,3,2].
+    If the element is odd, multiply it by 2.
+        For example, if the array is [1,2,3,4], then you can do this operation on the first element, and the array will be [2,2,3,4].
+
+  The deviation of the array is the maximum difference between any two elements in the array.
+
+Return the minimum deviation the array can have after performing some number of operations.
+  Ex. Input: nums = [1,2,3,4]
+      Output: 1
+      Explanation: You can transform the array to [1,2,3,2], then to [2,2,3,2], then the deviation will be 3 - 2 = 1.
+
+  Time  : O(N);
+  Space : O(N);
+*/
+
+class Solution {
+public:
+    int minimumDeviation(vector<int>& nums) {
+        priority_queue <int> pq;
+        int minimum = INT_MAX;
+        for(auto i : nums) {
+            if(i & 1)
+                i *= 2;
+            minimum = min(minimum, i);
+            pq.push(i);
+        }
+        int res = INT_MAX;
+        while(pq.top() % 2 == 0) {
+            int val = pq.top();
+            res = min(res, val - minimum);
+            minimum = min(val/2, minimum);
+            pq.pop();
+            pq.push(val/2);
+        }
+        return min(res, pq.top() - minimum);
+    }
+};