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

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Original file line number	Diff line number	Diff line change
`@@ -43,3 +43,59 @@ class Solution {`
`43`	`43`	`return (maxSum<0) ? res-1 : res;`
`44`	`44`	`}`
`45`	`45`	`}`
	`46`	`+`
	`47`	`+//Binary Search`
	`48`	`+`
	`49`	`+class Solution {`
	`50`	`+ public int maxValue(int n, int index, int maxSum) {`
	`51`	`+ //Binary Search code`
	`52`	`+ if(n == 1) return maxSum;`
	`53`	`+`
	`54`	`+ int left = 1, right = maxSum;`
	`55`	`+ while(left < right) {`
	`56`	`+ int mid = left + (right-left)/2;`
	`57`	`+ long sum = calculateSumOfArray(mid, index, n);`
	`58`	`+`
	`59`	`+ if(sum <= maxSum) {`
	`60`	`+ left = mid+1;`
	`61`	`+ }`
	`62`	`+ else {`
	`63`	`+ right = mid;`
	`64`	`+ }`
	`65`	`+ }`
	`66`	`+`
	`67`	`+ return left-1;`
	`68`	`+ }`
	`69`	`+`
	`70`	`+ //Calculate the Sum of array`
	`71`	`+ private long calculateSumOfArray(int v, int i, int n){`
	`72`	`+ long sum = 0;`
	`73`	`+ if(v <= i){`
	`74`	`+ int a = i+1-v;`
	`75`	`+ sum+=summation(v-1) + a;`
	`76`	`+ }`
	`77`	`+ else {`
	`78`	`+ int x = v-i;`
	`79`	`+ sum+= summation(v-1) - summation(x-1);`
	`80`	`+ }`
	`81`	`+`
	`82`	`+`
	`83`	`+ if(v <= n-i){`
	`84`	`+ int b = n-i-v;`
	`85`	`+ sum+=summation(v-1) + b;`
	`86`	`+ }`
	`87`	`+ else {`
	`88`	`+ int y = v-(n-i-1);`
	`89`	`+ sum+= summation(v-1) - summation(y-1);`
	`90`	`+ }`
	`91`	`+`
	`92`	`+ return sum+v;`
	`93`	`+`
	`94`	`+ }`
	`95`	`+`
	`96`	`+ //Utility method to apply summation formula`
	`97`	`+ private long summation(int n) {`
	`98`	`+ return (long)(n+1)*n/2;`
	`99`	`+ }`
	`100`	`+`
	`101`	`+}`

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