The naive brute force approach:
It is very simple to code but there are little things to look for (as least I have to) with respect to optimization. The outer loop runs for text, the inner loop runs for pattern with one more condition - **i+ len(pattern) <= len(text)**.
int bruteForce (text, pattern) for (i =0; i < len(text); i++):for (j = 0; j < len(pattern) && **i+ len(pattern) <= len(text)**; j++): //instead of i+j < len(text) if text[i+j] != pattern[j]: break; if (j==``len(pattern)``) return i; //match found return -1 //match not found
With i+j < len(text) (2+0 < 3), the inside loop runs this the last, even though not required i.e. i=2 is run.01**2** i aa**b** len(text)= 3 **a**x len(pattern)= 2 **0**1`` j
With **i+ len(pattern) <= len(text)** (1+2 <= 3), the inside loop runs this the last i.e. i=2 is not run.0**1**2 i a**a**b len(text)= 3 ax len(pattern)= **2** 01 j
Update: A simpler way is to run the outer loop for condition: i <= len(text) - len(pattern)