feat: solve NO.1717

1701-1800/1717. Maximum Score From Removing Substrings.md

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+# 1717. Maximum Score From Removing Substrings
+
+- Difficulty: Medium.
+- Related Topics: String, Stack, Greedy.
+- Similar Questions: Count Words Obtained After Adding a Letter.
+
+## Problem
+
+You are given a string `s` and two integers `x` and `y`. You can perform two types of operations any number of times.
+
+
+	Remove substring `"ab"` and gain `x` points.
+
+
+		
+- For example, when removing `"ab"` from `"cabxbae"` it becomes `"cxbae"`.
+
+
+	Remove substring `"ba"` and gain `y` points.
+
+		
+- For example, when removing `"ba"` from `"cabxbae"` it becomes `"cabxe"`.
+
+
+
+
+Return **the maximum points you can gain after applying the above operations on** `s`.
+
+
+Example 1:
+
+```
+Input: s = "cdbcbbaaabab", x = 4, y = 5
+Output: 19
+Explanation:
+- Remove the "ba" underlined in "cdbcbbaaabab". Now, s = "cdbcbbaaab" and 5 points are added to the score.
+- Remove the "ab" underlined in "cdbcbbaaab". Now, s = "cdbcbbaa" and 4 points are added to the score.
+- Remove the "ba" underlined in "cdbcbbaa". Now, s = "cdbcba" and 5 points are added to the score.
+- Remove the "ba" underlined in "cdbcba". Now, s = "cdbc" and 5 points are added to the score.
+Total score = 5 + 4 + 5 + 5 = 19.
+```
+
+Example 2:
+
+```
+Input: s = "aabbaaxybbaabb", x = 5, y = 4
+Output: 20
+```
+
+
+**Constraints:**
+
+
+
+- `1 <= s.length <= 105`
+
+- `1 <= x, y <= 104`
+
+- `s` consists of lowercase English letters.
+
+
+
+## Solution 1
+
+```javascript
+/**
+ * @param {string} s
+ * @param {number} x
+ * @param {number} y
+ * @return {number}
+ */
+var maximumGain = function(s, x, y) {
+    var stack = [];
+    var score = 0;
+    var a = x > y ? 'ab' : 'ba';
+    var b = x > y ? 'ba' : 'ab';
+    var ax = x > y ? x : y;
+    var bx = x > y ? y : x;
+    var score = 0;
+    for (var i = 0; i < s.length; i++) {
+        if (stack[stack.length - 1] === a[0] && s[i] === a[1]) {
+            stack.pop();
+            score += ax;
+        } else {
+            stack.push(s[i]);
+        }
+    }
+    s = stack.join('');
+    stack = [];
+    for (var i = 0; i < s.length; i++) {
+        if (stack[stack.length - 1] === b[0] && s[i] === b[1]) {
+            stack.pop();
+            score += bx;
+        } else {
+            stack.push(s[i]);
+        }
+    }
+    return score;
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n).
+* Space complexity : O(n).
+
+## Solution 2
+
+```javascript
+/**
+ * @param {string} s
+ * @param {number} x
+ * @param {number} y
+ * @return {number}
+ */
+var maximumGain = function(s, x, y) {
+    if (y > x) {
+        return maximumGain(s.split('').reverse(), y, x);
+    }
+    var aCount = 0;
+    var bCount = 0;
+    var score = 0;
+    for (var i = 0; i <= s.length; i++) {
+        if (s[i] === 'a') {
+            aCount += 1;
+        } else if (s[i] === 'b') {
+            if (aCount > 0) {
+                aCount -= 1;
+                score += x;
+            } else {
+                bCount += 1;
+            }
+        } else {
+            score += Math.min(aCount, bCount) * y;
+            aCount = 0;
+            bCount = 0;
+        }
+    }
+    return score;
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n).
+* Space complexity : O(1).