solution to problem 0528

dsa-problems/leetcode-problems/0500-0599.md

@@ -170,7 +170,7 @@ export const problems =[
     "problemName": "528. Random Pick with Weight",
     "difficulty": "Medium",
     "leetCodeLink": "https://leetcode.com/problems/random-pick-with-weight",
-    "solutionLink": "#"
+    "solutionLink": "/dsa-solutions/lc-solutions/0500-0599/0528-Random-Pick-with-Weight"
   },
   {
     "problemName": "529. Minesweeper",

dsa-solutions/lc-solutions/0500-0599/0528-Random-Pick-with-Weight.md

@@ -0,0 +1,153 @@
+---
+id: random-pick-with-weight
+title: Random Pick With Weight
+sidebar_label: 0528-Random-Pick-with-Weight
+tags:
+- Array
+- Math
+- Binary Search
+- Prefix Sum
+description: solution to the leetcode problem Random Pick With Weight.
+---
+
+## Problem
+
+You are given a 0-indexed array of positive integers `w` where `w[i]` describes the weight of the `i`th index.
+
+You need to implement the function `pickIndex()`, which randomly picks an index in the range `[0, w.length - 1]` (inclusive) and returns it. The probability of picking an index `i` is `w[i] / sum(w)`.
+
+For example, if `w = [1, 3]`, the probability of picking index `0` is `1 / (1 + 3) = 0.25` (i.e., 25%), and the probability of picking index `1` is `3 / (1 + 3) = 0.75` (i.e., 75%).
+
+Sure, here is the markdown with input, output, and explanation all in one go:
+
+### Examples
+
+**Example 1:**
+
+Input:
+```
+["Solution","pickIndex"]
+[[[1]],[]]
+```
+Output:
+```
+[null,0]
+```
+Explanation:
+```
+Solution solution = new Solution([1]);
+solution.pickIndex(); // return 0. The only option is to return 0 since there is only one element in w.
+```
+
+**Example 2:**
+
+Input:
+```
+["Solution","pickIndex","pickIndex","pickIndex","pickIndex","pickIndex"]
+[[[1,3]],[],[],[],[],[]]
+```
+Output:
+```
+[null,1,1,1,1,0]
+```
+Explanation:
+```
+Solution solution = new Solution([1, 3]);
+solution.pickIndex(); // return 1. It is returning the second element (index = 1) that has a probability of 3/4.
+solution.pickIndex(); // return 1
+solution.pickIndex(); // return 1
+solution.pickIndex(); // return 1
+solution.pickIndex(); // return 0. It is returning the first element (index = 0) that has a probability of 1/4.
+```
+
+Since this is a randomization problem, multiple answers are allowed.  
+All of the following outputs can be considered correct:
+```
+[null,1,1,1,1,0]
+[null,1,1,1,1,1]
+[null,1,1,1,0,0]
+[null,1,1,1,0,1]
+[null,1,0,1,0,0]
+......
+```
+and so on.
+
+### Constraints
+
+- $1 <= w.length <= 10^4$
+- $1 <= w[i] <= 10^5$
+- `pickIndex` will be called at most `10^4` times.
+```
+
+
+## Solution
+
+```cpp
+class Solution {
+  public:
+    Solution(vector<int>& w) {
+      int last = 0;
+      for (auto const& len : w) {
+        range.push_back({last, last + len});
+        last += len;
+      }
+      srand(time(nullptr)); 
+    }
+
+    int pickIndex() {
+      int logical_index = std::rand() % range.back().second; 
+      int lo = 0, hi = range.size() - 1;
+      while (lo <= hi) {
+        int mid = (lo + hi)/2;
+
+        if (logical_index >= range[mid].first && logical_index < range[mid].second) {
+          return mid;
+        } else if (logical_index < range[mid].first) {
+          hi = mid - 1;
+        } else {
+          lo = mid + 1;
+        }
+      }
+      return -1;
+    }
+  private:
+    vector<pair<int, int>> range;
+};
+```
+
+```java
+class Solution {
+
+    Random random;
+    int[] wSums;
+
+    public Solution(int[] w) {
+        this.random = new Random();
+        for(int i=1; i<w.length; ++i)
+            w[i] += w[i-1];
+        this.wSums = w;
+    }
+
+    public int pickIndex() {
+        int len = wSums.length;
+        int idx = random.nextInt(wSums[len-1]) + 1;
+        int left = 0, right = len - 1;
+        // search position 
+        while(left < right){
+            int mid = left + (right-left)/2;
+            if(wSums[mid] == idx)
+                return mid;
+            else if(wSums[mid] < idx)
+                left = mid + 1;
+            else
+                right = mid;
+        }
+        return left;
+    }
+}
+```
+
+### Complexity Analysis
+
+- **Time Complexity:** $O(\log N)$
+- **Space Complexity:** $O(N)$