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Create 1802. Maximum Value at a Given Index in a Bounded Array

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	`1`	`+//Bruteforce - O(Steps OR Result)/ O(1)`
	`2`	`+class Solution {`
	`3`	`+ public int maxValue(int n, int index, int maxSum) {`
	`4`	`+ int res = 1;`
	`5`	`+ maxSum -= n;`
	`6`	`+ int left = 0, right = 0;`
	`7`	`+ int maxLeft = index, maxRight = n - index - 1;`
	`8`	`+`
	`9`	`+ while(maxSum > 0) {`
	`10`	`+ res++;`
	`11`	`+ int leftVal = Math.min(left++, maxLeft);`
	`12`	`+ int rightVal = Math.min(right++, maxRight);`
	`13`	`+ maxSum -= (1 + leftVal + rightVal);`
	`14`	`+ }`
	`15`	`+ return (maxSum<0) ? res-1 : res;`
	`16`	`+ }`
	`17`	`+}`
	`18`	`+`
	`19`	`+`
	`20`	`+//Optimized - O(N)/O(1)`
	`21`	`+class Solution {`
	`22`	`+ public int maxValue(int n, int index, int maxSum) {`
	`23`	`+ int res = 1;`
	`24`	`+ maxSum -= n;`
	`25`	`+ int left = 0, right = 0;`
	`26`	`+ int maxLeft = index, maxRight = n - index - 1;`
	`27`	`+`
	`28`	`+ while(maxSum > 0) {`
	`29`	`+ res++;`
	`30`	`+ int leftVal = Math.min(left++, maxLeft);`
	`31`	`+ int rightVal = Math.min(right++, maxRight);`
	`32`	`+ maxSum -= (1 + leftVal + rightVal);`
	`33`	`+`
	`34`	`+ if(leftVal == maxLeft && rightVal == maxRight) {`
	`35`	`+ break;`
	`36`	`+ }`
	`37`	`+ }`
	`38`	`+`
	`39`	`+ if(maxSum > 0){`
	`40`	`+ res = res + (maxSum/n);`
	`41`	`+ }`
	`42`	`+`
	`43`	`+ return (maxSum<0) ? res-1 : res;`
	`44`	`+ }`
	`45`	`+}`

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