in the context of this question we can know that the starting a's and ending a's frequency is same for which the logic would be as same as a^n b^n as both this character has same frequency now for the middle character b the frequency of b depends on the value m and as it is at middle and after making sure we will have at least one b we will do the furthur process that is skipping b's and that is why pushing the initial a's then skipping the b's and poping a's will be the logic to design pda for this language the push pop and skip logic will be different for different problems
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love you brother thanks a lot
How we would know that which characters we have to push, which have to skip and which have to pop?
in the context of this question we can know that the starting a's and ending a's frequency is same for which the logic would be as same as a^n b^n as both this character has same frequency now for the middle character b the frequency of b depends on the value m and as it is at middle and after making sure we will have at least one b we will do the furthur process that is skipping b's and that is why pushing the initial a's then skipping the b's and poping a's will be the logic to design pda for this language the push pop and skip logic will be different for different problems
@Lucidd8 now I got it thanks!