# Math #### [7. Reverse Integer](https://leetcode.com/problems/reverse-integer/) Stop the loop before `res` exceeded the `limit`. ```java def reverse(self, x: int) -> int: limit = pow(2, 31) - 1 neg = x < 0 res = 0 x = abs(x) while x > 0: if res > limit / 10 and int(x % 10) != 0: return 0 res = res * 10 + int(x % 10) x = x // 10 if neg: return res * -1 return res ``` #### [12. Integer to Roman](https://leetcode.com/problems/integer-to-roman/) Store int value and its corresponding symbol values in two arrays. `While(num >= arr[i])` append result. ```java int[] arr = { 1,4,5,9,10,40,50,90,100,400, 500, 900, 1000}; String[] symbol = {"I", "IV", "V", "IX", "X", "XL", "L", "XC", "C", "CD", "D", "CM", "M"}; public String intToRoman(int num) { StringBuilder sb = new StringBuilder(); for (int i = arr.length - 1; i >= 0; i--) { while (num >= arr[i]) { sb.append(symbol[i]); num -= arr[i]; } } return sb.toString(); } ``` #### [13. Roman to Integer](https://leetcode.com/problems/roman-to-integer/) ```java public int romanToInt(String s) { Map map = new HashMap<>(); map.put('M', 1000); map.put('D', 500); map.put('C', 100); map.put('L', 50); map.put('X', 10); map.put('V', 5); map.put('I', 1); int res = 0; for (int i = 0; i < s.length(); i++) { if (i + 1 < s.length()) { if (map.get(s.charAt(i)) < map.get(s.charAt(i + 1))) { res += map.get(s.charAt(i + 1)) - map.get(s.charAt(i)); i++; } else { res += map.get(s.charAt(i)); } } else { res += map.get(s.charAt(i)); } } return res; } ``` #### [67. Add Binary](https://leetcode.com/problems/add-binary/) ```java public String addBinary(String a, String b) { int i = a.length() - 1; int j = b.length() - 1; StringBuilder sb = new StringBuilder(); int carry = 0; while (i >= 0 || j >= 0) { int one = i >= 0 ? a.charAt(i) - '0': 0; int two = j >= 0 ? b.charAt(j) - '0': 0; int sum = one + two + carry; sb.append(sum % 2); carry = sum / 2; i--; j--; } if (carry == 1) sb.append(1); return sb.reverse().toString(); } ``` #### [50. Pow(x, n)](https://leetcode.com/problems/powx-n/) 2^10 = 2\*2\*2\*2\*2\*2\*2\*2\*2\*2 (2\*2)\*(2\*2)\*(2\*2)\*(2\*2)\*(2\*2) 4^5 = (4\*4)\*(4\*4)\*4 16\*2 \* 4 {% tabs %} {% tab title="Java" %} ```java public double myPow(double x, int n) { if (n == 0) return 1.0; if (n == 1) return x; double res = 1.0; long longn = n; longn = longn < 0 ? -longn : longn; while (longn > 0) { if (longn % 2 != 0) { res *= x; } x *= x; longn /= 2; } return n < 0 ? 1.0 / res : res; } ``` {% endtab %} {% tab title="Python" %} ```python def myPow(self, x: float, n: int) -> float: negative = n < 0 i = abs(n) res = 1.0 while i > 0: if i % 2 != 0: res *= x x *= x i = i // 2 if negative: return 1.0 / res return res ``` {% endtab %} {% endtabs %} #### [273. Integer to English Words](https://leetcode.com/problems/integer-to-english-words/) ```java String[] LESS20 = {"", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"}; String[] TENs = {"", "Ten", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"}; String[] THOUSANDs = {"", "Thousand", "Million", "Billion"}; public String numberToWords(int num) { if (num == 0) return "Zero"; int i = 0; String res = ""; while (num != 0) { if (num % 1000 != 0) { res = helper(num % 1000) + THOUSANDs[i] + " " + res; } i++; num /= 1000; } return res.trim(); } private String helper(int n) { if (n == 0) return ""; if (n < 20) return LESS20[n] + " "; else if (n < 100) return TENs[n / 10] + " " + helper(n % 10); return LESS20[n / 100] + " Hundred " + helper(n % 100); } ``` #### 1131. Maximum of Absolute Value Expression Solve the equation by: ![](/files/-MFgLIX63xysiQM2gwg8) Then detect the maximum result of `max - min` ```java public int maxAbsValExpr(int[] arr1, int[] arr2) { int max1 = Integer.MIN_VALUE; int max2 = Integer.MIN_VALUE; int max3 = Integer.MIN_VALUE; int max4 = Integer.MIN_VALUE; int min1 = Integer.MAX_VALUE; int min2 = Integer.MAX_VALUE; int min3 = Integer.MAX_VALUE; int min4 = Integer.MAX_VALUE; for (int i = 0; i < arr1.length; i++) { max1 = Math.max(max1, arr1[i] + arr2[i] + i); max2 = Math.max(max2, arr1[i] + arr2[i] - i); max3 = Math.max(max3, arr1[i] - arr2[i] - i); max4 = Math.max(max4, arr1[i] - arr2[i] + i); min1 = Math.min(min1, arr1[i] + arr2[i] + i); min2 = Math.min(min2, arr1[i] + arr2[i] - i); min3 = Math.min(min3, arr1[i] - arr2[i] - i); min4 = Math.min(min4, arr1[i] - arr2[i] + i); } return Math.max(Math.max(max1 - min1, max2 - min2), Math.max(max3 - min3, max4 - min4)); } ``` #### [470. Implement Rand10() Using Rand7()](https://leetcode.com/problems/implement-rand10-using-rand7/) [How to do question like implement RandomX by RandomY.](https://leetcode.com/problems/implement-rand10-using-rand7/discuss/150301/Three-line-Java-solution-the-idea-can-be-generalized-to-%22Implement-RandM\(\)-Using-RandN\(\)%22) ![](/files/-MMMg4bDLGVFu5tjAakH) ```java public int rand10() { int res = 40; while (res >= 40) { res = (rand7() - 1) * 7 + rand7() - 1; } return res % 10 + 1; } ``` [2918. Minimum Equal Sum of Two Arrays After Replacing Zeros](https://leetcode.com/problems/minimum-equal-sum-of-two-arrays-after-replacing-zeros/) ```typescript function minSum(nums1: number[], nums2: number[]): number { let sum1 = 0, countZero1 = 0; for (const num of nums1) { sum1 += num; if (num === 0) { countZero1++; sum1++; } } let sum2 = 0, countZero2 = 0; for (const num of nums2) { sum2 += num; if (num === 0) { countZero2++; sum2++; } } if ((countZero1 === 0 && sum2 > sum1) || (countZero2 === 0 && sum1 > sum2)) return -1; return Math.max(sum1, sum2); }; ``` [3335. Total Characters in String After Transformations I](https://leetcode.com/problems/total-characters-in-string-after-transformations-i/) ```typescript function lengthAfterTransformations(s: string, t: number): number { const MOD = 1_000_000_007; const freq: number[] = new Array(26).fill(0); for (const c of s) { freq[c.charCodeAt(0) - 'a'.charCodeAt(0)]++; } let zIdx = 25; while (t-- > 0) { const next = (zIdx + 1) % 26; freq[next] = (freq[next] + freq[zIdx]) % MOD; zIdx = (zIdx - 1 + 26) % 26; } let result = 0; for (const f of freq) { result = (result + f) % MOD; } return result; }; ``` --- # Agent Instructions: Querying This Documentation If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question. 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