Create count-the-number-of-substrings-with-dominant-ones.cpp

Author: kamyu104

C++/count-the-number-of-substrings-with-dominant-ones.cpp

@@ -0,0 +1,64 @@
+// Time:  O(n * sqrt(n)) = O(n^(3/2))
+// Space: O(n)
+
+// two pointers, sliding window
+class Solution {
+public:
+    int numberOfSubstrings(string s) {
+        int result = 0;
+        vector<int> idxs = {-1};
+        for (int i = 0; i < size(s); ++i) {
+            if (s[i] == '0') {
+                idxs.emplace_back(i);
+            }
+        }
+        idxs.emplace_back(size(s));
+        for (int c = 0; c * c <= size(s); ++c) {
+            for (int i = 0, left = 1, right = 1, cnt = 0; i < size(s); ++i) {
+                if (idxs[right] == i) {
+                    ++right;
+                } else {
+                    ++cnt;
+                }
+                if (right - left == c + 1) {
+                    cnt -= (idxs[left] - 1) - idxs[left - 1];
+                    ++left;
+                }
+                if (!(right - left == c && cnt >= c * c)) {
+                    continue;
+                }
+                result += min((c ? idxs[left] : i) - idxs[left - 1], cnt - c * c + 1);
+            }
+        }
+        return result;
+    }
+};
+
+// Time:  O(n * sqrt(n)) = O(n^(3/2))
+// Space: O(1)
+// two pointers, sliding window
+class Solution2 {
+public:
+    int numberOfSubstrings(string s) {
+        int result = 0;
+        for (int c = 0; c * c <= size(s); ++c) {
+            vector<int> cnt(2);
+            for (int right = 0, left = 0, curr = 0; right < size(s); ++right) {
+                ++cnt[s[right] == '1'];
+                while (cnt[0] == c + 1) {
+                    --cnt[s[left++] == '1'];
+                }
+                if (!(cnt[0] == c && cnt[1] >= c * c)) {
+                    continue;
+                }
+                for (curr = max(curr, min(left, right)); curr < right; ++curr) {
+                    if (s[curr] == '0') {
+                        break;
+                    }
+                }
+                result += min(curr - left + 1, cnt[1] - c * c + 1);              
+            }
+        }
+        return result;
+    }
+};