Thus, in both the string the frequency of each letter must be the same if they are an anagram of each other. Thus our algorithm checks and compare frequency of each letter in both the strings.The strings to be anagram of each other, their length must be same. Let n1 be the length ...
Thus, one of the digits in the string 𝑎1𝑎2𝑎3…𝑎𝑛−1a1a2a3…an−1 is e too big; i.e., 𝑎𝑘=𝑐+𝑒ak=c+e for some 1≤𝑘≤𝑛−11≤k≤n−1 and where c is the correct digit. Now, the single error applied to (2) yields 𝑎1𝑃+𝑎2𝑃2+…...
Thus, one of the digits in the string 𝑎1𝑎2𝑎3…𝑎𝑛−1a1a2a3…an−1 is e too big; i.e., 𝑎𝑘=𝑐+𝑒ak=c+e for some 1≤𝑘≤𝑛−11≤k≤n−1 and where c is the correct digit. Now, the single error applied to (2) yields 𝑎1𝑃+𝑎2𝑃2+…...