713. Subarray Product Less Than K | leetcode daily challenge | DSA | Shashwat Hindi

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Problem Name:
713. Subarray Product Less Than K

Problem Statement:
Given an array of integers nums and an integer k, return the number of contiguous subarrays where the product of all the elements in the subarray is strictly less than k.

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Timestamp:
0:00 - Introduction

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those who wanna check Brute Force Approach :--> (Bhaiya, please pin this comment)
class Solution {
public int nums, int k) {
int n = nums.length; // length of the array
int ans = 0;
for(int i=0; i<n; i++){
int product = nums[i];
if(product < k) ans++;
for(int j=i+1; j<n; j++){
product *= nums[j];
if(product < k) ans++;
if(product >= k) break; // if you not use this line, it fine
}
}
return ans;
}
}

Inspire_with_SM

Thanks for easy & detailed explaination 🙏🏼

AbhijeetMuneshwar

Go for 2 pointer approach, where 1st pointer (i) would be fixed and second (j) moves on and multiplies the elements at position (j). If that product is less than K, count++ otherwise stop moving second pointer and start again from i by increasing i++;

utpalkant

bro i got the solution but i am not able to understand like how the sliding window is working exactly like how would it cover all the subarrays, should i know sliding window concept first for it to understand ?

kanaklata

bhaiya please ek handwritten notes bana dijiye OOPs ke topic pr mey padh to leta hu but at the end bahut confusion rheta hai OOP mey plz ek handwritten notes.

Inspire_with_SM

Bhaiya, yeh program bhi sahi h naa?

public class Solution {
public int nums, int k) {
// Variable Sliding Window
int len = nums.length;
int i = 0, j = 0;
long prod = 1;
int cnt = 0;

while (j < len) {
prod *= nums[j];
if (prod < k) {
cnt++;
j++;
if (j >= len) {
i++;
j = i;
prod = 1;
}
} else {
i++;
j = i;
prod = 1;
}
}
return cnt;
}
}

imPriyansh

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