Add C++ solution of problem #3152

Author: Cuda-Chen

Diff for: 3152/solution.cpp

@@ -0,0 +1,63 @@
+// use binary search to the start in interval then check whether end is in the same interval of start or not
+
+class Solution {
+public:
+    vector<bool> isArraySpecial(vector<int>& A, vector<vector<int>>& queries) {
+        int n = A.size();
+
+        vector<bool> ans(queries.size(), false);
+        int idx = 0; // start index of special
+        int cur = A[0];
+        vector<int> start, end; // start and end of specials
+        int i;
+        for(i = 1; i < n; i++) {
+            if(
+                (cur % 2 == 0 && A[i] % 2 == 1) ||
+                (cur % 2 == 1 && A[i] % 2 == 0)
+            ) {
+                // switch
+                cur = A[i];
+                continue;
+            }
+
+            start.push_back(idx);
+            end.push_back(i - 1);
+            idx = i;
+            cur = A[i];
+        }
+        // deal with leftover
+        if(i - idx > 1) {
+            start.push_back(idx);
+            end.push_back(i - 1);
+        }
+
+        for(i = 0; i < queries.size(); i++) {
+            int s = queries[i][0];
+            int e = queries[i][1];
+            int ii = b_search(start, s);
+
+            if(
+                (s == e) ||
+                (ii < start.size() && e <= end[ii])
+            )
+                ans[i] = true;
+        }
+
+        return ans;
+    }
+
+private:
+    // Return the index of less than or equal to target.
+    int b_search(vector<int> &A, int target) {
+        int l = 0;
+        int r = A.size();
+        while(l < r) {
+            int m = l + (r - l) / 2;
+            if(A[m] <= target)
+                l = m + 1;
+            else
+                r = m;
+        }
+        return l - 1;
+    }
+};