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0 <= goal <= nums.length

22:T4c0,# Intuition\nWe will use PrefixSum algorithm and HashMap to solve this problem\n# Approach\nwe count prefix sum and then we iterate over each prefix sum and check that prefix sum are goal more than current prefix sum, if true count it.\n# Complexity\n- Time complexity:\nO(n) -> we iterate 2n times\n- Space complexity:\nO(n) -> we use 2n to store prefix sum and map to count prefix sums\n# Code\n```\nfunction numSubarraysWithSum(nums: number[], goal: number): number {\n // declare prefixSum array and fill it with 0\n const prefix = new Array(nums.length + 1).fill(0);\n // iterate over nums array:\n for(let i = 0; i < nums.length; i++) {\n // calculate prefix sum on each step\n prefix[i + 1] = prefix[i] + nums[i];\n }\n // declare map to count prefix sums\n const map = new Map();\n // declare result counter\n let count = 0;\n // iterate over each prefix sum\n prefix.forEach((sum) => {\n // count if prefix sum are in map\n count += map.has(sum) ? map.get(sum) : 0;\n // count prefix sum that are at least goal more than current prefix sum\n map.set(sum + goal, map.has(sum + goal) ? map.get(sum + goal) + 1 : 1)\n })\n // return count\n return count\n};\n```23:T710,**Brute Force Solution**\r\n# Complexity\r\n- Time complexity: O(N^2)\r\n\r\n\r\n- Space complexity: O(1)\r\n\r\n```\r\n\r\nvar numSubarraysWithSum = function(nums, goal) {\r\n \r\n let totalSubarray = 0;\r\n\r\n for(let i=0; i goal) break\r\n }\r\n }\r\n\r\n return totalSubarray\r\n};\r\n\r\n```\r\n**Prefix Sum**\r\n# Complexity\r\n- Time complexity: O(N)\r\

Binary Subarrays With Sum

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