### Problem

Given an array of n positive integers and a positive integer s, find the minimal length of a subarray of which the sum ≥ s. If there isn't one, return 0 instead.

### Example

Given the array [2,3,1,2,4,3] and s = 7, the subarray [4,3] has the minimal length of 2 under the problem constraint.

### JavaScript Code

function minSubArrayLen(s, nums) { if(nums == null || nums.length == 0){ return 0; } // initialize min length to be the input array length var result = nums.length; var i = 0; var sum = nums[0]; for(var j=0; j<nums.length; ){ if(i==j){ if(nums[i]>=s){ return 1; //if single elem is large enough }else{ j++; if(j<nums.length){ sum = sum + nums[j]; }else{ return result; } } }else{ //if sum is large enough, move left cursor if(sum >= s){ result = Math.min(j-i+1, result); sum = sum - nums[i]; i++; //if sum is not large enough, move right cursor }else{ j++; if(j<nums.length){ sum = sum + nums[j]; }else{ if(i==0){ return 0; }else{ return result; } } } } } return result; } console.log(minSubArrayLen(7, [2,3,1,2,4,3]));

this my another solution. first you have to sort thee array and then return with the specify position which you want.

ReplyDeletevar arr=[3,2,1,5,6,4];

function nthHighest(arr, n) {

var sorted = arr.sort(function (a, b) {

return a - b;

});

return sorted[sorted.length - n];

}

console.log(nthHighest(arr,n));

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