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	`1`	`+/**`
	`2`	`+Given an array nums of integers, we need to find the maximum possible sum of elements of the array such that it is divisible by three.`
	`3`	`+`
	`4`	`+`
	`5`	`+`
	`6`	`+Example 1:`
	`7`	`+`
	`8`	`+Input: nums = [3,6,5,1,8]`
	`9`	`+Output: 18`
	`10`	`+Explanation: Pick numbers 3, 6, 1 and 8 their sum is 18 (maximum sum divisible by 3).`
	`11`	`+Example 2:`
	`12`	`+`
	`13`	`+Input: nums = [4]`
	`14`	`+Output: 0`
	`15`	`+Explanation: Since 4 is not divisible by 3, do not pick any number.`
	`16`	`+`
	`17`	`+*/`
	`18`	`+package main`
	`19`	`+`
	`20`	`+funcmaxSumDivThree(nums []int)int {`
	`21`	`+dp:=make([]int,3)`
	`22`	`+for_,v:=rangenums {`
	`23`	`+tmp:=make([]int,3)`
	`24`	`+copy(tmp,dp)`
	`25`	`+fori:=0;i<3;i++ {`
	`26`	`+dp[(v+tmp[i])%3]=max(dp[(v+tmp[i])%3],v+tmp[i])`
	`27`	`+}`
	`28`	`+}`
	`29`	`+returndp[0]`
	`30`	`+}`
	`31`	`+`
	`32`	`+funcmax(x,yint)int {`
	`33`	`+ifx>y {`
	`34`	`+returnx`
	`35`	`+}`
	`36`	`+returny`
	`37`	`+}`
	`38`	`+`
	`39`	`+/**`
	`40`	`+dp:除以3的余数有3种情况，0，1，2`
	`41`	`+*/`

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	`1`	`+/**`
	`2`	`+Given an array arr, replace every element in that array with the greatest element among the elements to its right, and replace the last element with -1.`
	`3`	`+`
	`4`	`+After doing so, return the array.`
	`5`	`+`
	`6`	`+`
	`7`	`+`
	`8`	`+Example 1:`
	`9`	`+`
	`10`	`+Input: arr = [17,18,5,4,6,1]`
	`11`	`+Output: [18,6,6,6,1,-1]`
	`12`	`+`
	`13`	`+*/`
	`14`	`+package main`
	`15`	`+`
	`16`	`+funcreplaceElements(arr []int) []int {`
	`17`	`+varres []int=make([]int,len(arr))`
	`18`	`+varmaxint=arr[len(arr)-1]`
	`19`	`+res[len(arr)-1]=-1`
	`20`	`+fori:=len(arr)-2;i>=0;i-- {`
	`21`	`+res[i]=max`
	`22`	`+ifarr[i]>max {`
	`23`	`+max=arr[i]`
	`24`	`+}`
	`25`	`+}`
	`26`	`+returnres`
	`27`	`+}`
	`28`	`+`
	`29`	`+/**`
	`30`	`+trick:from right to left`
	`31`	`+*/`

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