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Create Max Sum of a Pair With Equal Sum of Digits

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	`1`	`+想法:因為可能的 digitsum 介於 1-81 之間,所以可以用一個矩陣紀錄每個 digitsum 目前看到的最大數值`
	`2`	`+如果前面有出現過的話則考慮更新答案,或是出現比之前更大的數值的時候更新最大數值`
	`3`	`+`
	`4`	`+// assume there are n elements , the max element value is m`
	`5`	`+Time Complexity : O(n * logm) for traversing all elements and each element takes O(logm) to find out its digit sum`
	`6`	`+Space Complexity : O(logm) for there are O(logm) possible digit sums`
	`7`	`+`
	`8`	`+class Solution {`
	`9`	`+public:`
	`10`	`+ int calculatedigitsum(int num) {`
	`11`	`+ int sum = 0 ;`
	`12`	`+ while (num > 0) {`
	`13`	`+ sum += num % 10 ;`
	`14`	`+ num /= 10 ;`
	`15`	`+ }`
	`16`	`+ return sum ;`
	`17`	`+ }`
	`18`	`+ int maximumSum(vector<int>& nums) {`
	`19`	`+ int ans = -1 ;`
	`20`	`+ vector<int> digitmax(82) ;`
	`21`	`+`
	`22`	`+ for(auto &i : nums) {`
	`23`	`+ int digitsum = calculatedigitsum(i) ;`
	`24`	`+`
	`25`	`+ if (digitmax[digitsum] != 0)`
	`26`	`+ ans = max(ans , digitmax[digitsum] + i) ;`
	`27`	`+ digitmax[digitsum] = max(digitmax[digitsum] , i) ;`
	`28`	`+ }`
	`29`	`+`
	`30`	`+`
	`31`	`+ return ans ;`
	`32`	`+ }`
	`33`	`+};`

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