Solution as on 19-02-2022 06:30 am

Author: knockcat

1675. Minimize Deviation in Array.cpp

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+// 1675.✅ Minimize Deviation in Array
+
+class Solution
+{
+public:
+    int minimumDeviation(vector<int> &nums)
+    {
+        int n = nums.size();
+        int mx = INT_MIN, mn = INT_MAX;
+
+        // Increasing all elements to as maximum as it can and tranck the minimum,
+        // number and also the resutl
+        for (int i = 0; i < n; ++i)
+        {
+            if ((nums[i] % 2) != 0) // multiplication by 2 if nums[i] is odd
+                nums[i] *= 2;       // maximising all odd numbers
+
+            mx = max(mx, nums[i]);
+            mn = min(mn, nums[i]);
+        }
+
+        int min_deviation = mx - mn;
+
+        priority_queue<int> pq;
+        // Inserting into Priority queue (Max Heap) and try to decrease as much we can
+        for (int i = 0; i < n; ++i)
+        {
+            pq.push(nums[i]);
+        }
+
+        while ((pq.top()) % 2 == 0)
+        {
+            int top = pq.top();
+            pq.pop(); // popped the top element
+
+            min_deviation = min(min_deviation, top - mn);
+            top /= 2;
+            mn = min(mn, top); // updating min
+            pq.push(top);      // pushing again the top as we have to minimize the max
+        }
+
+        min_deviation = min(min_deviation, pq.top() - mn);
+
+        // we are returning mx - mn
+
+        return min_deviation;
+    }
+};
+
+class Solution
+{
+public:
+    int minimumDeviation(vector<int> &nums)
+    {
+        set<int> s;
+
+        // Storing all  elements  in sorted order
+        // insert even directly and odd with one time multiplication
+        // and it will become even
+        for (int i = 0; i < nums.size(); ++i)
+        {
+            if (nums[i] % 2 == 0)
+                s.insert(nums[i]);
+
+            else
+                // Odd number are transformed
+                // using 2nd operation
+                s.insert(nums[i] * 2);
+        }
+
+        // maximum - minimun
+        int diff = *s.rbegin() - *s.begin();
+
+        // run the loop untill difference is minimized
+        while (*s.rbegin() % 2 == 0)
+        {
+
+            // Maximum element of the set
+            int x = *s.rbegin();
+            s.erase(x);
+            // remove begin element and inserted half of it for minimizing
+            s.insert(x / 2);
+
+            diff = min(diff, *s.rbegin() - *s.begin());
+        }
+        return diff;
+    }
+};