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	`1`	`+/**`
	`2`	`+Given an integer array A, you partition the array into (contiguous) subarrays of length at most K. After partitioning, each subarray has their values changed to become the maximum value of that subarray.`
	`3`	`+`
	`4`	`+Return the largest sum of the given array after partitioning.`
	`5`	`+`
	`6`	`+`
	`7`	`+`
	`8`	`+Example 1:`
	`9`	`+`
	`10`	`+Input: A = [1,15,7,9,2,5,10], K = 3`
	`11`	`+Output: 84`
	`12`	`+Explanation: A becomes [15,15,15,9,10,10,10]`
	`13`	`+`
	`14`	`+*/`
	`15`	`+`
	`16`	`+package main`
	`17`	`+`
	`18`	`+funcmaxSumAfterPartitioning(A []int,Kint)int {`
	`19`	`+n:=len(A)`
	`20`	`+dp:=make([]int,n+1)`
	`21`	`+fori:=0;i<n;i++ {`
	`22`	`+curMax:=A[i]`
	`23`	`+forj:=1;j<=K&&i-j+1>=0;j++ {`
	`24`	`+curMax=max(curMax,A[i-j+1])`
	`25`	`+dp[i+1]=max(dp[i+1],dp[i-j+1]+curMax*j)`
	`26`	`+}`
	`27`	`+}`
	`28`	`+returndp[n]`
	`29`	`+}`
	`30`	`+`
	`31`	`+funcmax(a,bint)int {`
	`32`	`+ifa>b {`
	`33`	`+returna`
	`34`	`+}`
	`35`	`+returnb`
	`36`	`+}`
	`37`	`+`
	`38`	`+/**`
	`39`	`+note:DP`
	`40`	`+*/`

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	`1`	`+/**`
	`2`	`+For strings S and T, we say "T divides S" if and only if S = T + ... + T (T concatenated with itself 1 or more times)`
	`3`	`+`
	`4`	`+Return the largest string X such that X divides str1 and X divides str2.`
	`5`	`+`
	`6`	`+`
	`7`	`+`
	`8`	`+Example 1:`
	`9`	`+`
	`10`	`+Input: str1 = "ABCABC", str2 = "ABC"`
	`11`	`+Output: "ABC"`
	`12`	`+Example 2:`
	`13`	`+`
	`14`	`+Input: str1 = "ABABAB", str2 = "ABAB"`
	`15`	`+Output: "AB"`
	`16`	`+Example 3:`
	`17`	`+`
	`18`	`+Input: str1 = "LEET", str2 = "CODE"`
	`19`	`+Output: ""`
	`20`	`+`
	`21`	`+*/`
	`22`	`+package main`
	`23`	`+`
	`24`	`+import"golang.org/x/tools/go/ssa/interp/testdata/src/strings"`
	`25`	`+`
	`26`	`+funcgcdOfStrings(str1string,str2string)string {`
	`27`	`+`
	`28`	`+iflen(str1)==0\|\|str1==str2 {`
	`29`	`+returnstr1`
	`30`	`+}`
	`31`	`+iflen(str2)==0 {`
	`32`	`+returnstr2`
	`33`	`+}`
	`34`	`+iflen(str1)>len(str2) {`
	`35`	`+x:=strings.Index(str1,str2)`
	`36`	`+ifx>-1 {`
	`37`	`+returngcdOfStrings(str2,str1[x+len(str2):])`
	`38`	`+}`
	`39`	`+}else {`
	`40`	`+y:=strings.Index(str2,str1)`
	`41`	`+ify>-1 {`
	`42`	`+returngcdOfStrings(str1,str2[y+len(str1):])`
	`43`	`+}`
	`44`	`+}`
	`45`	`+return""`
	`46`	`+}`
	`47`	`+`
	`48`	`+/**`
	`49`	`+need to know the function:Strings.Index`
	`50`	`+*/`

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	`1`	`+/**`
	`2`	`+Given the root of a binary tree, consider all root to leaf paths: paths from the root to any leaf. (A leaf is a node with no children.)`
	`3`	`+`
	`4`	`+A node is insufficient if every such root to leaf path intersecting this node has sum strictly less than limit.`
	`5`	`+`
	`6`	`+Delete all insufficient nodes simultaneously, and return the root of the resulting binary tree.`
	`7`	`+`
	`8`	`+*/`
	`9`	`+package main`
	`10`	`+`
	`11`	`+typeTreeNodestruct {`
	`12`	`+Valint`
	`13`	`+Left*TreeNode`
	`14`	`+Right*TreeNode`
	`15`	`+}`
	`16`	`+`
	`17`	`+funcsufficientSubset(rootTreeNode,limitint)TreeNode {`
	`18`	`+sufficientSubsetRec(&root,0,limit)`
	`19`	`+returnroot`
	`20`	`+}`
	`21`	`+`
	`22`	`+funcsufficientSubsetRec(ptrToCurr**TreeNode,accint,limitint) {`
	`23`	`+cur:=*ptrToCurr`
	`24`	`+valueSoFar:=acc+cur.Val`
	`25`	`+`
	`26`	`+hasLeftKid,hasRightKid:=cur.Left!=nil,cur.Right!=nil`
	`27`	`+ifcur.Left!=nil {`
	`28`	`+sufficientSubsetRec(&cur.Left,valueSoFar,limit)`
	`29`	`+}`
	`30`	`+ifcur.Right!=nil {`
	`31`	`+sufficientSubsetRec(&cur.Right,valueSoFar,limit)`
	`32`	`+}`
	`33`	`+ifvalueSoFar<limit&&cur.Left==nil&&cur.Right==nil {`
	`34`	`+*ptrToCurr=nil`
	`35`	`+}`
	`36`	`+ifcur.Left==nil&&cur.Right==nil&& (hasLeftKid\|\|hasRightKid) {`
	`37`	`+*ptrToCurr=nil`
	`38`	`+}`
	`39`	`+}`

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